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UNIVERSITY OF CALIFORNIA AT BERKELEY
DEPARTMENT OF MECHANICAL ENGINEERING
ME 185
P. M. NAGHDI’S NOTES ON CONTINUUM MECHANICS
1994
UNIVERSITY OF CALIFORNIADepartment of Mechanical Engineering
ME185
A List of General Referencesiiiiiiiiiiiiiiiiiiiiiiii
1. Cartesian Tensorsiiiiiiiiiiiiiii, by Sir Harold Jeffreys, Cambridge UniversityPress, 1952 [lst ed., 1931, 7th impression 1969].
2. Schaum’s Outline Series - Theory and Problems of ContinuumMechanics, by G. E. Mase, McGraw-Hill Book Co., 1970.
3. Chapters 1-7 of Mechanicsiiiiiiiii, by E. A. Fox, Harper & Row, 1967.
4. Continuum Mechanicsiiiiiiiiiiiiiiiiiii, by Y. C. Fung, Prentice-Hall Inc., 1969.
5. Continuum Mechanicsiiiiiiiiiiiiiiiiiii, by P. G. Hodge, Jr., McGraw-Hill, 1970.
6. Mechanics of Continuaiiiiiiiiiiiiiiiiiii, by A. C. Eringen, John Wiley & Sons,Inc., 1967.
7. Continuum Mechanicsiiiiiiiiiiiiiiiiiii, by P. Chadwick, G. Allen & UnwinLtd., London, 1976.
8. Continuum Mechanicsiiiiiiiiiiiiiiiiiii, by A.J.M. Spencer, Longman GroupLtd., London, 1980.
9. An Introduction to the Theory of Elasticityiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii, by R. J. Atkin andN. Fox, Longman Inc., New York, 1980.
Table of Contents
I. Kinematics .......................................................................................................................... 11. Bodies, configurations, motions, mass, and mass density. ..................................... 12. Deformation gradient. Measure of strain. Rotation and stretch. ............................ 133. Further developments of kinematical results. ......................................................... 204. Further developments of kinematical results. ......................................................... 205. Velocity gradient. Rate of deformation. Vorticity. ................................................. 216. Superposed rigid body motions. ............................................................................. 257. Infinitesimal deformation and infinitesimal strain measures. ................................. 358. The transport theorem. ............................................................................................ 41
II. Conservation Laws and Some Related Results ................................................................. 441. Conservation of mass. .............................................................................................. 45
Appendix to Sec. 1 of Part II. .................................................................................. 472. Forces and couples. Euler’s laws. .......................................................................... 493. Further consideration of the stress vector. Existence of the stress tensor andits
relationship to the stress vector. ............................................................................. 514. Derivation of spatial (or Eulerian) form of the equations of motion. ..................... 585. Derivation of the equations of motion in referential (or Lagrangian form). .......... 616. Invariance under superposed rigid body motions. .................................................. 647. The principle of balance of energy. ........................................................................ 71
III. Examples of Constitutive Equations and Applications. Inviscid and Viscous Fluidand Nonlinear and Linear Elasticity. .......................................................................... 741. Inviscid fluid. .......................................................................................................... 752. Viscous fluid. .......................................................................................................... 793. Elastic solids: Nonlinear constitutive equations. .................................................... 834. Elastic solids: Linear constitutive equations. ......................................................... 865. Some steady flows of viscous fluids. ...................................................................... 916. Some unsteady flow problems. ............................................................................... 967. Further constitutive results in linear elasticity and illustrative examples. .............. 988. Saint-Venant torsion of a circular cylinder. ............................................................ 102Appendix to Part III (Secs. 5-8). .................................................................................. 104
Supplements to the Main Text ................................................................................................ 110Appendix L: Some results from linear algebra ...................................................................... 150
ME185
Part I: Kinematics
1. Bodies, configurations, motions, mass, and mass density.
A body is a set of particles. Sometimes in the literature a body B is referred to as a mani-
fold of particles. Particles of a body B will be designated as X (see Fig. 1.1) and may be
identified by any convenient system of labels, such as for example a set of colors. In mechanics
the body is assumed to be smooth and, by assumption, can be put into correspondence with a
domain of Euclidean space. Thus, by assumption, a particle X can be put into a one-to-one
correspondence with the triples of real numbers X1,X2,X3 in a region of Euclidean 3-space E3.
The mapping from the body manifold onto the domain of E3 is assumed to be one-to-one, inver-
tible, and differentiable as many times as desired; and for most purposes two or three times
suffice.
Each part (or subset) S of the body B at each instant of time is assumed to be endowed with
a positive measure M(S), i.e., a real number > 0, called the mass of the part S and the whole body
is assumed to be endowed with a nonnegative measure M(B) called the mass of the body. We
return to a further consideration of mass later in this section.
Bodies are seen only in their configurations, i.e., the regions of E3 they occupy at each
instant of time t (−∞ < t < +∞). These configurations should not be confused with the bodies
themselves.
Consider a configuration of the body B at time t in which configuration B occupies a region
R in E3 bounded by a closed surface ∂R (Fig. 1.2). Let x be the position vector of the place
occupied by a typical particle X at time t. Since the body can be mapped smoothly onto a
domain of E3, we write
x = χh
(X,t) . (1.1)
In (1.1), X refers to the particle, t is the time, x [the value of the function χh
] is the place occupied
by the particle X at time t and the mapping function χh
is assumed to be differentiable as many
times as desired both with respect to X and t. Also, for each t, (1.1) is assumed to be invertible
so that
- 2 -
X = χh
−1 (x,t) , (1.2)
where the symbol χ−1 designates the inverse mapping. The mapping (1.1) is called a motion of
the body B. The description of motion in (1.1) is similar to that in particle mechanics and is
traditionally referred to as the material description.
During a motion of the body B, a particle X (by occupying successive points in E3) moves
through space and describes a path C ; the equation of this path is parametrically represented by
(1.1). The rate of change of place with time as X traverses C is called the velocity and is tangent
to the curve C. Similarly, the second rate of change of place with time as X traverses C is the
acceleration. Thus, in the material description of the motion, the velocity v and acceleration a
are defined by
v = x.
=∂t
∂χh
(X,t)hhhhhhh , a = x..
=∂t2
∂2χh
(X,t)hhhhhhhh . (1.3)
In the above formulae, the particle velocity v is the partial derivative of the function χh
with
respect to t holding X fixed; and, similarly, the particle acceleration a is the second partial
derivative of χh
with respect to t holding X fixed. Also, the superposed dot, which designates the
material time differentiation, can be utilized in conjunction with any function f and signifies dif-
ferentiation with respect to t holding X fixed.
Reference configuration. Often it is convenient to select one particular configuration and
refer everything concerning the body and its motion to this configuration. The reference
configuration need not necessarily be a configuration that is actually occupied by the body in any
of its motions. In particular, the reference configuration need not be the initial configuration of
the body.
Let κR be a reference configuration of B in which X is the position vector of the place occu-
pied by the particle X (Fig. 1.3). Then, the mapping from X to the place X in the configuration
κR may be written as
- 3 -
X = κR (X) (1.4)
which specifies the position vector X occupied by the particle X in the reference configuration
κR. The mapping (1.4) is assumed to be differentiable as many times as desired and invertible.
The inverse mapping is
X = κR
−1(X) . (1.5)
Then, the mapping (1.1) at time t from the place x to the particle X may be expressed in terms of
(1.5), i.e.,
x = χh
[κR
−1(X),t] = χ
hκR
(X,t) . (1.6)
The second of (1.6), which involves the function χh
κR(with a subscript κR, emphasizes that the
motion described in this manner represents a sequence of mappings of the reference
configuration κR. In the future, however, for simplicity’s sake we omit the subscript κR and
write (1.6)2 and its inverse as
x = χ(X,t) , X = χ−1(x,t) (1.7)
with the understanding that X in the argument (1.6) is the position vector of X in the reference
configuration κR. Returning to (1.6)2, we observe that for each κR a different function χh
κR
results; and the choice of reference configuration, similarly to the choice of coordinates, is arbi-
trary and is introduced for convenience.
A necessary and sufficient condition for the invertibility of the mappings of (1.7)1,2 is that
the determinant J of the transformation from X to x be nonzero, i.e.,
J = det(∂X
∂χhhhh) = det(∂XA
∂χihhhhh) ≠ 0 . (1.8)
- 4 -
Description of the motion. There are several methods of describing the motion of a body.
Here, we describe three but note that, because of our smoothness assumptions, they are all
equivalent.
It was noted earlier that the description utilized in (1.1) is the material description. In such
a description one deals with abstract particles X which together with time t are the independent
variables.
A description such as (1.7) in which the position X of X at some time, e.g., t = 0, is used as
a label for the particle X is called the referential description. In the referential description,
which is also known as Lagrangian, the independent variables are X and t. For some purposes,
it is convenient to display and to utilize (1.7) in terms of its rectangular Cartesian coordinates.
Thus, let EK be constant orthonormal basis vectors associated with the reference configuration
and similarly denote by ek constant orthonormal basis vectors associated with the present
configuration at time t. Then, the positions X and x referred to EK and ek, respectively, are
X = XK EK , x = xk ek (1.9)
where XK are the rectangular Cartesian coordinates of the position X and similarly xk are the
rectangular Cartesian coordinates of the position x. Referred to the basis ek, the component
form of (1.7)1 can be displayed as
xk = χk(XK,t) , (1.10)
where without loss in generality (since EK are constant orthonormal basis) we have also replaced
the argument X by its components XK. In the future we often display the various formulae both
in their "coordinate-free" forms, as well as their component forms.
The particle velocity and acceleration in a material description of motion are given, respec-
tively, by (1.3)1,2. The corresponding expressions for velocity and acceleration in the referential
description (1.7) will have the same forms as (1.3)1,2 but with the argument X replaced by X:
- 5 -
v = x.
=∂t
∂χhhh(X,t) , a = x..
=∂t2
∂2 χhhhh(X,t) . (1.11)
So far we have introduced two descriptions of motion. The third is called the spatial
description in which attention is centered in the present configuration, i.e., the region of space
occupied by the body at the present time t. In the spatial description, which is also known as
Eulerian, the independent variables are the place x and the time t. To elaborate further, consider
an entity f defined by
f = f(X,t) . (1.12)
Since (1.1) is invertible in the form (1.2), the function f(X,t) can be expressed as a different
function of x,t, i.e.,
f(X,t) = f(χh
−1(x,t),t) = f(x,t) . (1.13)
Moreover, the function f is unique. It is clear that with the use of the transformation of the form
(1.13), the velocity and acceleration in (1.3)1,2 can be expressed in terms of different functions of
x,t:
v = v(X,t) = v(x,t) = vk ek , a = a(X,t) = a(x,t) = ak ek . (1.14)
In the above spatial description, the spatial form f(x,t) on the right-hand side of (1.13) was
obtained from a representation (1.12) in which f is a function of particle X and t. Analogous
results hold if we start with f = f(X,t) and then use (1.7)2 to obtain
f = f(X,t) = f(x,t) . (1.15)
- 6 -
Material derivative. Consider again a function such as f(X,t) in (1.12). The partial
derivative of this function with respect to t holding the variable X fixed, denoted by f., is called
the material derivative of f. But f, with the use of (1.13)2, can also be expressed as a different
function of x,t. Thus, using the chain rule for differentiation, we have
f.
=∂t∂fhhh c
c X
=∂x∂fhhh .
∂t
∂χh
hhh +∂t∂fhhh , (1.16)
where on the left-hand side of the above we have temporarily emphasized that the variable X is
held fixed when calculating the partial derivative of f with respect to t. With the use of the first
of (1.3)1, we may rewrite (1.16)2 as
f.
=∂t
∂f(X,t)hhhhhhh cc X
=∂t∂fhhh +
∂x∂fhhh . v =
∂t∂fhhh + vk ∂xk
∂fhhhh (1.17)
which is the material derivative of f(x,t). The results of the type (1.15) and (1.17) are applicable
to any scalar-valued, vector-valued or tensor-valued field. In particular, consider the referential
form of the velocity vector v which by (1.11)1 and transformation of the type (1.15) is
v = v(X,t) = v(x,t) . (1.18)
Then, the acceleration or the material derivative of v where these are regarded as functions of x,t
are
a = v.
=∂t∂vhhh +
∂x∂vhhh v = a(x,t) . (1.19)
Material curve, material surface and material volume. Let a body B with material
points (or particles) X in a fixed reference configuration κR occupy a region R R in E3 bounded
by a closed surface ∂R R. Any subset SR of B in E3 will be designated by PR (⊆ R R) bounded
by a closed boundary surface ∂PR. Because of the nature of the smoothness assumption
- 7 -
imposed on the mapping (1.7)1 from the reference configuration κR to the current configuration κ
of B at time t, all curves, surfaces and volumes in κR are carried by the motion χ into curves, sur-
faces and volumes in the configuration κ. Thus, the region R R with the closed boundary surface
∂R R is mapped into a region R with corresponding closed boundary surface ∂R in κ. More-
over, the parts PR (⊆ R R) bounded by ∂PR is mapped into a part P (⊆ R) bounded by a closed
surface ∂P. The part P is occupied by the same material points (or particles) as those which
occupied PR and such a region or volume is called a material volume.
A surface in the reference configuration κR is defined by equations of the form
F(X) = 0 or X = X(U1,U2) , (1.20)
where U1,U2 are parametric variables on the surface. With the use of (1.7)2, the surface (1.20)
can be mapped into a corresponding surface
f(x,t) = F(χ−1(x,t)) = 0 or x(U1,U2,t) = χ[X(U1,U2),t] (1.21)
in the configuration κ at time t. The surface (1.21) is called a material surface, since its material
points (or particles) are the same as those of the surface (1.20).
A curve in the reference configuration may be regarded as the intersection of two surfaces
of the form
F(X) = 0 , G(X) = 0 , (1.22)
and is defined by
X = X(U) , (1.23)
where U is a parameter and can be identified with the arc length. The intersection of the surfaces
(1.22) can be mapped into a curve in the configuration κ which is the intersection of the surfaces
f(x,t) = F(χ−1(x,t)) = 0 , g(x,t) = G(χ−1(x,t)) = 0 (1.24)
- 8 -
or
x(U,t) = χ[X(U),t] . (1.25)
The curve (1.25) is called a material curve, since its material points (particles) are the same as
those of the curve (1.23).
Lagrange’s criterion for a material surface. A surface f(x,t) = 0 is said to be material if
and only if the material derivative of f vanishes, i.e.,
f.
=∂t∂fhhh +
∂x∂fhhh . v = 0 . (1.26)
First we write f(x,t) = F(X,t) with the use of (1.7)1 and suppose that f(x,t) is material.
Then, F(X,t) must be independent of t and hence F.
= 0 and this implies that f.
= 0.
Next, suppose that f.
= 0. Then, it follows that F.
= 0 and this implies that F must be
independent of t and a function of X only so that F(X) = 0 and after using (1.7)2 we have the
result that F(χ−1(x,t)) = f(x,t) = 0 is material.
In a similar manner, we may establish that the intersection of surfaces f(x,t) = 0 and
g(x,t) = 0 is a material curve if and only if
f.
=∂t∂fhhh +
∂x∂fhhh . v = 0 , g
.=
∂t∂ghhh +
∂x∂ghhh . v = 0 . (1.27)
Mass density. Before closing this section, consider again a configuration κ of the body B
at time t in which B occupies a region of E3 R bounded by a closed surface ∂R and recall that a
part (or a subset) of B will be denoted by S. The part S in a neighborhood of the place x in κ
occupies a region of space P (⊆ R) bounded by a closed surface ∂P. We introduce explicitly the
mass M(S) of the part and the mass M(B) of the whole body at time t. These measures are non-
negative and may be functions of t. Assuming that the measure M(S) is absolutely continuous,
- 9 -
then the limit
ρ =v→0lim
v
M(St)hhhhh (1.28)
exists, where v = v(P) is the volume of the region of space P. The scalar field ρ = ρ(x,t) is
called the mass density of the body in the configuration at time t. The mass density is nonnega-
tive and is a function of both x and t. The mass of the part S of the body and the mass of the
whole body, both at time t, can be expressed in terms of the mass density ρ:
M(St) = ∫P
ρ dv , M(Bt) = ∫R
ρ dv , (1.29)
where the subscript t attached to S and B emphasizes that the left-hand sides of (1.29)1,2 refer to
the masses of S and B in the configuration κ at time t and dv is an element of volume in this
configuration, the regions of integration being over P and R in E3. Similarly, we define the mass
density ρR
in a reference configuration and write
M(SR) = ∫P
R
ρR dVR , M(BR) = ∫R
R
ρR dVR , (1.30)
where dVR denotes an element of volume in the reference configuration and PR,R R (PR ⊆ R R)
are the regions of space occupied by the part and the whole body in the reference configuration.
If the reference configuration is identified with the initial configuration at time to, then instead of
(1.30) we write
M(So) = ∫Po
ρo dV , M(Bo) = ∫R o
ρo dV , (1.31)
where ρo and dV are, respectively, the mass density and the volume element in the initial
configuration while Po and R o (Po ⊆ R o) are the regions of space occupied, respectively, by the
part and the whole body in the initial configuation.
It should be emphasized that, at this stage in our development, the mass of a part of the
body and that of the whole body depend on the particular configuration occupied by the body.
- 10 -
2. Deformation gradient. Measures of strain. Rotation and stretch.
Henceforth we identify the particle (material point) X by its position vector X in a fixed
reference configuration, which (for convenience) we take to be coincident with the initial
configuration κo. The deformation gradient (also called displacement gradient) F relative to the
reference position is defined by
F = F(X,t) =∂X
∂χ(X,t)hhhhhhh , J = det F ≠ 0 . (2.1)
The components of F are designated by FiA or sometimes written as xi,A with a comma designat-
ing partial differentiation. They are given by
FiA =∂XA
∂χi(XB,t)hhhhhhhh (2.2)
and may be regarded as a 3×3 square matrix.
At a fixed time t, line elements dx are related to line elements dX in κo by
dx = F dX or dxi = xi,A dXA . (2.3)
The deformation gradient F in (2.3) describes the local deformation of a particle (material point)
whose position vector is X in κo. In other words, (2.3) is a linear transformation of a small
neighborhood of X from the reference configuration κo into the current configuration κ at time t.
Let the magnitudes of dX and dx be denoted by dS and ds, respectively; and let M and m
be the unit vectors in the directions of dX and dx, respectively. Then,
dX = MdS , M . M = 1 ,
(2.4)
dx = m ds , m . m = 1 ,
or in component forms
- 11 -
dXA = MA dS , MA MA = 1 ,
(2.5)
dxi = mi ds , mi mi = 1 .
Measures of strain.
In general, the material line element dX undergoes both stretch and rotation. The ratio ds/dS ,
denoted by λ, is called the stretch of the line element, i.e.,
λ =dSdshhh . (2.6)
It then follows from (2.3)-(2.5) that
m ds = dx = F dX = F M dS ,
(2.7)
or mi ds = dxi = xi,A dXA = xi,A MA dS .
Next, using (2.6), from (2.7) we have
λm = F M or λmi = xi,A MA . (2.8)
By taking the scalar product of each side of (2.8) with itself we obtain
λ2 = F M . F M = M . C M , (2.9)
where in writing (2.9)3 we have also defined the tensor C by
C = FT F . (2.10)
The component forms of the equations (2.9)-(2.10) are:
λ2 = CAB MA MB , (2.9a)
and
- 12 -
CAB = xi,A xi,B . (2.10a)
The symmetric tensor C (with components CAB) is called the right Cauchy-Green tensor.
Clearly, for a material line element which coincides with dX in the reference configuration
κo, the value of the stretch λ in the direction M can be calculated from (2.9) once C is known.
Alternatively, λ can be calculated in terms of the deformed line element dx in the current
configuration κ. To see this, since F is invertible, we rewrite (2.3) as
F−1 dx = dX or XA,i dxi = dXA , (2.11)
where the inverse function F−1 using (1.7)2 is defined by
F−1 =∂x
∂χ−1hhhhh or XA,i =
∂xi
∂χA
−1
hhhhh . (2.12)
Then, from (2.4), (2.5), (2.11) and (2.12) we have
M dS = dX = F−1 dx = F−1 m ds ,
(2.13)
or MA dS = dXA = XA,i dxi = XA,i mi ds ,
which with the use of (2.6) can be written in the form
λ−1 M = F−1 m or λ−1 MA = XA,i mi . (2.14)
The result (2.14) can also be obtained immediately by inverting (2.8). By taking the scalar pro-
duct of each side of (2.14) with itself we obtain
λ−2 = F−1 m . F−1 m = m . (F−1)T F−1 m = m . B−1 m , (2.15)
where in writing (2.15)3 we have also defined the tensor B by
- 13 -
B = F FT . (2.16)
The component forms of equations (2.15)-(2.16) are:
λ−2 = cij mi mi , (2.15a)
cij = bij
−1= XA,i XA,j , bij = xi,A xj,A , (2.16a)
where the notation cij designates the components of the inverse of the tensor B. The symmetric
tensor B whose components are given in (2.16a) is called the left Cauchy-Green tensor. The ten-
sors C and B represent, respectively, the referential (or Lagrangian) and the spatial (or Eulerian)
descriptions of strain. When the motion is rigid, both tensors become identity tensors, i.e.,
C = B = I, where I is the unit tensor.
For some purposes, it is convenient to employ relative measures of strain such that these
measures vanish when the motion is rigid. Recalling (2.9) and (2.15) we may write
λ2 − 1 = M . (C − I)M = (CAB − δAB)MA MB
(2.17)
1 − λ−2 = m . (I − B−1) m = (δij − bij
−1) mi mj
or equivalently,
λ2 − 1 = 2M . E M = 2EAB MA MB ,
(2.18)
1 − λ−2 = 2m . Ehh
m = 2eij mi mj ,
where
E =21hh(C − I) or EAB =
21hh(CAB − δAB) , (2.19)
Ehh
=21hh( I − B−1) or eij =
21hh(δij − cij) . (2.20)
- 14 -
The relative measure E defined by (2.19) is known as the relative Lagrangian strain, and the
corresponding relative spatial measure of strain Ehh
is defined by (2.20).
Rotation and stretch.
The deformation gradient F in (2.3) describes the local deformation of a material line element at
X from the reference configuration κo to the current configuration κ; it involves, in general, both
rotation and stretch. Now, since by (1.8) the deformation gradient F is nonsingular, using the
polar decomposition theorem it may be expressed in the polar forms
F = R U = V R or xi,A = RiB UBA = Vij RjA , (2.21)
where U and V are positive definite symmetric tensors called, respectively, the right and left
stretch tensors and R is the proper orthogonal tensor satisfying
R RT = RT R = I det R = 1 , (2.22)
or in component form
RiA RiB = δAB or RiA RjA = δij . (2.22a)
The effect of (2.21) is to replace the linear transformation (2.3) by two linear transformations:
Either by
dX′ = U dX and dx = R dX′ or dXB′ = UBA dXA and dxi = RiB dXB
′ , (2.23)
or by
dx′ = R dX and dx = V dx′ or dxj′ = RjA dXA and dxi = Vij dxj
′ . (2.24)
Associated with each of the stretch tensors U, V are three positive principal values (eigen-
values) and three orthogonal principal directions (eigenvectors). These will be discussed later.
- 15 -
Here, it should be emphasized that, in general, F, R, U, V are all functions of X and t and they
can vary from one material point to another during motion. In the remainder of this section, we
provide physical interpretations for the decompositions (2.23) and (2.24).
(i) Stretch followed by pure rotation. In this case, let the deformation of dX into dX′
involve only pure stretch and the deformation from dX′ into dx be one of pure rotation. By
(2.21)1 and (2.23), as well as (2.9)-(2.10), we have
dx . dx = R dX′ . R dX′ = dX′ . dX′ , (2.25a)
or
dxi dxi = RiA RiB dXA′ dXB
′ = δAB dXA′ dXB
′ = dXA′ dXA
′ , (2.25b)
and
C = FT F = (R U)T (R U) = U2 , (2.26)
or
CAB = xi,A xi,B = RiK UKA RiL ULB = δKL UKA ULB = UKA UKB , (2.27a)
λ2 = UKA UKB MA MB = CAB MA MB . (2.27b)
According to (2.25), the length of the line element dX′ is the same as the length of dx so that all
the stretching is represented by U in (2.26). However, line elements are generally also rotated
by the action of U. The part of the deformation described by R is a pure rotation.
(ii) Pure rotation followed by stretch. In this case, let the deformation of dX into dx′ be
one of pure rotation and the deformation from dx′ to dx involve only stretch and rotation. By
(2.21) and (2.24), as well as (2.15)-(2.16), we have
dx′ . dx′ = R dX . R dX = RT R dX . dX = dX . dX , (2.28)
- 16 -
or
dxi′ dxi
′ = RiA RiB dXA dXB = δAB dXA dXB = dXA dXA (2.29)
and
B = F FT = (V R) (V R)T = V2 , (2.30a)
λ2
1hhh = m . B−1 m , (2.30b)
or
bij = xi,A xj,A = Vir RrA Vjs RsA = Vir Vjr (2.31a)
λ−2 = (B−1)ij mi mj . (2.31b)
According to (2.28), the length of the line element dx′ is the same as the length of dX so that the
deformation described by R is one of pure rotation.
Keeping in mind the decomposition (2.21), it is evident from (2.26) and (2.30a) that
U = (FT F)1⁄2 and V = (F FT)
1⁄2 involve the square root operation which often is difficult to exe-
cute and this is the main reason for the choice of U2 = C and V2 = B as the strain tensors.
Moreover, since U2 is symmetric positive definite, it possesses a unique symmetric positive
definite square root; a parallel remark applies to the square root of V2. In this regard the expres-
sions for the square root of the stretch λ and its inverse, in the forms (2.9) and (2.15), are espe-
cially noteworthy.
- 17 -
Secs. 3 & 4. Further developments of kinematical results.
- 18 -
5. Velocity gradient. Rate of deformation. Vorticity.
Consider the spatial (Eulerian) form of the particle velocity in the form given by (1.14),
v = v(x,t) and define the spatial velocity gradient tensor by
L = grad v(x,t) (5.1)
or
∂x∂vhhh = vi,j ei ×b ej , vi,j =
∂xj
∂vihhhh , (5.2)
where vi,j are the components of L referred to the orthonormal basis ei and in recording (5.2) we
have used the same symbol for the function v and its value. The tensor L can be uniquely
decomposed into its symmetric part D and its skew-symmetric part W (see Appendix L, Sect. 8)
as follows:
L = D + W , (5.3)
where
D =21hh(L + LT) = dij ei ×b ej , W =
21hh(L − LT) = wij ei ×b ej . (5.4)
The corresponding component forms of (5.3) and (5.4) are:
vi,j = dij + wij , (5.3a)
where
dij =21hh(vi,j + vj,i) = dji , wij =
21hh(vi,j − vj,i) = − wji . (5.4a)
The tensor D with components dij is called the rate of deformation tensor and W with com-
ponents wij is called the vorticity or the spin tensor.
There is a one-to-one correspondence between a skew tensor and an axial vector (see
Appendix L, Sect. 12). The axial vector w which corresponds to the skew vorticity tensor W in
(5.3) or (5.3a) is defined by
W z = w × z or w =21hh curl v (5.5)
- 19 -
for every vector z. Referred to the basis ei, w reads as
w = wi ei (5.6)
and the corresponding component forms of (5.5)1,2 are:
wij = − εijk wk = εjik wk , wi =21hh εijk wkj =
21hh εijk vk,j , (5.5a)
where εijk stand for the components of the permutation symbol. Motion in which w = 0 is
called irrotational motion. It is usual in some books to identify curl v (instead of21hh curl v in
(5.5)2) as vorticity and denote it by w = curl v.
The foregoing results have been obtained starting from (5.1), where v = v(x,t). Alterna-
tively, we may begin by recalling the referential (Lagrangian) form of the particle velocity in the
form given by (1.14)1, where v = v(X,t). Then, the material derivative of the deformation gra-
dient (2.1) is
F.
= L F , (5.7)
or in component form
xi,A
hhh.= x
.i,A = vi,A = vi,k xk,A . (5.7a)
Additional useful results may be obtained from (5.7) or (5.7a). Thus, the material derivative of
the Jacobian of transformation (1.8) is
J.
= det Fhhhh.
= J tr (L F F−1) = J tr L
= J tr D = J vk,k = J dkk (5.8)
or
J.
= J XA,i vi,k xk,A = J δik vi,k = J vk,k . (5.8a)
A motion which is volume-preserving, i.e., a motion corresponding to which the volume occu-
pied by any material region remains unchanged, is called isochoric. For an isochoric motion,
- 20 -
J = 1 , J.
= 0 => tr D = dkk = div v = vk,k = 0 . (5.9)
Next, consider the material derivative of the tensor C defined by (2.10):
C.
= FTFhhhh.
= F.
T F + FT F.
= FT( LT + L)F = 2FT D F , (5.10)
or in component form
C.
AB = xi,A xi,B
hhhhhh.= xi,A
hhh.xi,B + xi,A xi,B
hhh.
= vi,k xk,A xi,B + vi,k xk,B xi,A
= (vi,k + vk,i)xk,A xi,B = 2dik xk,A xi,B (5.10a)
In order to discuss a physical interpretation of the tensors D and W (or equivalently the
axial vector w), we take the material derivative of (2.8) with respect to t and make use of (5.7) to
obtain
λ mhhhh.
= λ.
m + λ m.
= F Mhhhh.
= L F M = λ L m , (5.11)
or in component form
λ mi
hhhh.= λ
.mi + λ m
.i = xi,A MA
hhhhhhh.= vi,k xk,A MA = λ vi,k mk . (5.11a)
The scalar product of (5.11) with the unit vector m, after also using the results m . m = 1,
m . m.
= 0, yields
λλ.hh = l n λ
hhhh.= L m . m = (D + W) . m ×b m = D . m ×b m , (5.12)
or
λλ.hh = l n λ
hhhh.= vi,k mi mk = (dik + wik)mi mk = dik mi mk . (5.12a)
- 21 -
It follows from (5.12) that the rate of deformation D determines the material time derivative of
the logarithmic stretch for a material line element having the direction m in the current
cofiguration κ. Further, using (5.12), the result (5.11)2 may be rewritten as
m.
= L m −λλ.hh m = W m + [D − (
λλ.hh) I]m , (5.13)
or from (5.11a)
m.
i = vi,j mj − (λλ.hh)mi = wij mj + [dij − (
λλ.hh)δij ] mj . (5.13a)
Now, let m be a principal direction of the eigenvectors associated with the rate of deformation
tensor D. Then,
D m = γ m or dij mj = γ mi (5.14)
where the associated scalar eigenvalue γ is
γ = D m . m = D . m ×b m = dij mi mj =λλ.hh (by (5.12a)) (5.15)
where we have also used (5.12). It follows from (5.14) and (5.15) that
D m =λλ.hh m or dij mj =
λλ.hh mi . (5.16)
Moreover, from combination of (5.13) and (5.16) we deduce that
m.
= W m = w × m or m.
i = wij mj = εikj wk mj . (5.17)
Thus, the material time derivative of a unit vector m (which represents the direction of a
material line in κ) in a principal direction of D is determined by (5.17). In other words, the axial
vector w (or the vorticity) is the angular velocity of the line element which is parallel to a princi-
pal direction.
(For further results, see the Supplement to Part I, Section 5.)
- 22 -
6. Superposed rigid body motions.
A rigid motion is one that preserves relative distance. In this section, we examine the effect
of such a motion on the various kinematic measures introduced in preceding sections. More
specifically, we consider here motions which differ from a prescribed motion, such as (1.7)1,
only by superposed rigid body motions of the whole body, i.e., motions which in addition to a
prescribed motion involve purely rigid motions of the body. For later reference, we recall here
the motion defined in Eq. (1.7)1, namely
x = χ (X, t) .
Consider a material point (or particle) of the body, which in the present configuration at
time t occupies the place x as specified by (1.7)1. Suppose that under a superposed rigid body
motion, the particle which is at x at time t moves to a place x+ at time t+ = t + a, a being a con-
stant. In what follows for all quantities associated with the superposed motion we use the same
symbols as those associated with the motion (1.7)1 to which we also attach a plus "+" sign.
Thus, we introduce a vector function χ+ and write
x+ = χ+ (X, t+) = χ+ (X, t) . (6.1)
It is clear that the difference between the two functions χ+ and χ+ is due to the presence of the
constant a in the argument of the former.
Similarly, consider another material point of the body which in the present configuration at
time t occupies the place y specified by
y = χ (Y, t) . (6.2)
Suppose that under the same superposed rigid body motion, the particle which is at y at time t
moves to a place y+ at time t+. Then, corresponding to (6.1), we write
y+ = χ+ (Y, t+) = χ+ (Y, t) (6.3)
Recalling the inverse relationship X = χ−1(x, t) and the analogous result for Y = χ−1(y, t), the
function χ+ on the right-hand sides (6.1) and (6.3) may be expressed as different functions of x, t
- 23 -
and y, t, respectively, i.e.,
x+ = χ+(X, t) = χh
+(x, t) , y+ = χ+(Y, t) = χh
+(y, t) . (6.4)
During the superposed rigid body motions of the whole body, the magnitude of the relative dis-
placement χh
+(x, t) - χh
+(y, t) must remain unaltered for all pairs of material points X, Y in the
body and for all t in some finite time interval [t1, t2]. Hence,
[χh
+(x, t) − χh
+(y, t)] . [χh
+(x, t) − χh
+(y, t)] = (x − y) . (x − y) (6.5)
for all x, y in the region R occupied by the body at time t.
Since x, y are independent, we may differentiate (6.5) successively with respect to x and y
to obtain the following differential equation for χh
+:
RJQ ∂x
∂χh
+(x, t)hhhhhhhh
HJP
TRJQ ∂y
∂χh
+(y, t)hhhhhhhh
HJP
= I , (6.6)
where x, y are any points in R and t is any time in the interval [t1, t2]. Since the tensor ∂χh
+(y,
t)/∂y is invertible, (6.6) can be written in the alternative form
RJQ ∂x
∂χh
+(x, t)hhhhhhhhHJP
T
=RJQ ∂y
∂χh
+(y, t)hhhhhhhh
HJP
−1
= QT(t) (6.7)
for all x, y ⊆ R and t in [t1, t2]. Thus, each side of the equation (6.7) is a tensor function of t, say
QT(t), and we may set
∂x
∂χh
+(x, t)hhhhhhhh = Q(t) for x ∈ R . (6.8)
Therefore we also have∂y
∂χh
+(y, t)hhhhhhhh = Q(t). >From (6.6) or (6.7), we conclude that
QTQ = I , det Q = ± 1 , (6.9)
and hence Q is an orthogonal tensor. The motion under consideration must include the particu-
lar case χh
+(x, t) = x and for this particular case Q = I and det Q = +1. Since the motions are con-
tinuous, we must alwaysiiiiii have
- 24 -
det Q = 1 (6.10)
and therefore Q(t) is a proper orthogonal tensor.
The differential equation (6.8) may be integrated to yield
χh
+(x, t) = a(t) + Q(t)x , (6.11)
where a(t) is a vector function of time. The last result is a general solution of (6.5) for χh
+(x, t).
For later convenience, we put
a(t) = c+(t+) − Q(t)c(t) , (6.12)
where c+, c are vector functions of t+ and t, respectively. With the use of (6.12), the (6.11) can
be expressed in the form
x+ = c+ + Q(x − c) , (6.13)
where
QT Q = Q QT = I , det Q = 1 . (6.14)
The transformation (6.13) is a rigid transformation since it is distance preserving, i.e.,
| x+ − y+ | 2 = (x+ − y+) . (x+ − y+) = Q(x − y) . Q(x − y)
= (x − y) . [QTQ(x − y)]
= (x − y) . I(x − y)
= (x − y) . (x − y) = | x − y | 2 . (6.15)
The transformation (6.13) also preserves the angle between any two nonzero vectors x − y and
x − z since
cosθ+ =| x+z+ |
x+ − y+hhhhhhh
=| x − y |
Q(x − y)hhhhhhhh . Q(x
=| x − y | | x − z |
(x − y) . [QTQ(x − z)]hhhhhhhhhhhhhhhhhhh
=| x − y |(x − y)hhhhhhh .
| x − z |(x − z)hhhhhhh = cosθ , (6.16)
- 25 -
where z is any place in the configuration occupied by the body at time t which moves to z+ at
time t+ as a consequence of the rigid body motion. Since (6.13) is both length and angle preserv-
ing, it follows that the element of length, element of area and element of volume all remain unal-
tered under superposed rigid body motions.
The component forms of the various results between (6.5) and (6.10) proceed as follows.
First, the component form of (6.5) is
[χh
i+(xj, t) − χ
hi+(yj, t)][χ
hi+(xj, t) − χ
hi+(yj, t)] (6.5a)
To record the results (6.6) in component form, we differentiate (6.5a) successively with respect
to xm and yn and obtain
∂xmj
∂χh
i+(xj t)hhhhhhhh , t)] = δim(xi − yi)
and
∂xm
∂χh
i+(xj, t)hhhhhhhhh
∂yn
∂χh
i(yj, t)hhhhhhhh (6.6a)
Next, multiply (6.6a) by∂χh
k+(yj, t)
∂ynhhhhhhhhh and use the chain rule for differentiation, i.e.,
∂yn
∂χh
i+(yj, t)hhhhhhhhh
∂χh
k+(yj, t)
∂ynhhhhhhhhh
to deduce
∂xm
∂χh
k+(xj, t)hhhhhhhhh =
∂χh
k+(yj, t)
∂ymhhhhhhhhh . (6.7a)
Since the left-hand side of (6.7a) is independent of yj and the right-hand side is independent of
xj, each side must be a function of time only and we have
∂xm
∂χh
k+(xj, t)hhhhhhhhh = Qkm(t) . (6.8a)
- 26 -
Integration of the differential equation (6.8a) yields
xk+ = χ
hk+(xj, t) = ak(t) (6.11a)
Since xj is any point, we must also conclude from (6.11a) that
yk+ = χ
hk(yj, t) = ak(t) + Qkm(t)ym
and hence
∂yn
∂χh
k+(yj, t)hhhhhhhhh = Qkn(t) .
Substitution of the last result and (6.8a) into (6.6a) results in the component for of
Qkm(t)Qkn(t) = δmn . (6.9a)
Also, for the convenience of the reader, we also record here the component forms of most of the
equations given above. Thus, for example, the component form of (6.1) is
xi+ = χi
+(XA, t+) = χi+(XA,t) . (6.1a)
Similarly, we identify the component form of (6.2), (6.3), etc., as
yi = χi(YA, t) , (6.2a)
yi+ = χi
+(YA, t+) = χi+(YA, t) , (6.3a)
xi+ = χi
+(XA, t) = χh
i+(xj, t) , yi
+ = χi+(YA, t) = χ
hi+(yj, t) , (6.4a)
[χh
i+(xj, t) − χ
hi+(yj, t)][χ
hi+(xj, t) − χ
hi+(yj, t)] = (xi − yi)(xi − yi) , (6.5a)
∂xm
∂χh
i+(xj, t)hhhhhhhhh
∂yn
∂χh
i+(yj, t)hhhhhhhhh = δmn , (6.6a)
∂xk
∂χh
i+(xj, t)hhhhhhhhh = Qik(t) , (6.8a)
Qim(t)Qin(t) = δmn , (6.9a)
- 27 -
χh
i+(xj, t) = ai(t) + Qik(t)xk , (6.11a)
ai(t) = ci+(t+) − Qik(t)ck(t) , (6.12a)
xi+ = ci
+ + Qij(xj − cj) , (6.13a)
Qim Qin = δmn = Qmi Qni , (6.14a)
(xi+ − yi
+)(xi+ − yi
+)k)
= δjk(xj − yj)(xk − yk) = (xj − yj)(xj − yj) , (6.15a)
| x+ − y+ | | x+ − z+ | cosθ+i+)
= Qim(xm − ym)Qin(xn − zn)
= δmn(xm − ym)(xn − zn)
= (xm − ym)(xm − zm)
= | x − y | | x − z | cosθ . (6.17a)
Recall
x+ = χ+(X, t+) = χ+(X, t) = χh
+(x, t) ,
or
x+ = χh
+(x, t+) = a(t) + Q(t)x . (6.11)
Then,
v+ = x. + =
dt+dx+hhhh =
∂t+∂χ+(X, t+)hhhhhhhhh
=∂t
∂χ+(X, t)hhhhhhhh
dt+dthhh
=∂t
∂χ+(X, t)hhhhhhhh (6.18)
Therefore
v+ = x. + =
dt+dhhh(a(t) + Q(t)x)
- 28 -
=dtdhhh(a(t) + Q(t)x)
dt+dthhh
= a.(t) + Q
.(t)x + Q(t)x
.
= a.(t) + Q
.(t)x + Q(t)v . (6.19)
Let
Ω (t) = Q.
(t)QT(t) . (6.20)
Then
Ω Q = Q.
QT Q = Q.
. (6.21)
But, from (6.9) QT Q = I, so that
Q.
T(t)Q(t) + QT(t)Q.
(t) = O ,
[Ω (t)Q(t)]TQ(t) + QT(t)Ω(t)Q(t) = O ,
QT(t)ΩT(t)Q(t) + QT(t)Ω(t)Q(t) = O ,
ΩT(t) + Ω(t) = O
or
ΩT(t) = −Ω(t) . (6.22)
Since Ω (t) is skew-symmetric, it posseses an axial vector
ωi = −21hh εijk Ωjk , (6.23)
or
Ωjk = −εjki ωi .
Alternatively, since Ω is skew-symmetric, there exists a vector-valued function ω such that for
any vector V,
- 29 -
Ω V = ω × V . (6.24)
Returing to (6.19), we may write
v+ = a.
+ Ω Q x + Q v . (6.25)
- 30 -
So far we have been discussing superposediiiiiiiii rigid body motions. In order to relate our results to
the more familiar equations of rigid body dynamics, let us now consider a rigid motion defined
by
x = X , (6.26)
x+ = a(t) + Q(t)x= a(t) + Q(t)X . (6.27)
The equation (6.27)1 may be solved for x in the form
x = QT(t)(x+ − a(t)) . (6.28)
Then
v+ = a.(t) + Q
.(t)x
= a.(t) + Ω Q(t)[QT(t)(x+
= a.(t) + Ω (t)[x+ − a(t)]
= a.(t) + ωωωω(t) × [x+ − a(t)] . (6.29)
Thus ωωωω(t) is recognized to be the angular velocity of the body.
The component form of the foregoing equations are listed below:
xi+ = ai(t) + Qij(t)xj , (6.11a)
vi+ = x
.i+ = a
.i(t) + Q
.ij(t)xj + Qij(t)x
.j , (6.19a)
Ωik(t) = Q.
im(t)Qkm(t) , (6.20a)
ΩikQkl = Q.
imQkmQkl = Q.
il , (6.21a)
Recalling
Q.
il Qjl + Qil Q.
jl = 0 ,
- 31 -
we obtain
Ωij(t) = −Ωji(t) , (6.22a)
ΩimVm = −εimp ωp Vm = εipm ωp Vm , (6.24a)
vi+ = a
.i + Ωim Qmn xn + Qim vm , (6.25a)
xi+ = ai(t) + Qim xm , (6.27a)
Qij xi+ = Qij ai + Qij Qim xm
Qij(xi+ − ai) = δjm xm = xj , (6.28a)
vi+ = a
.i + Ωij(xj
+ − aj)
= a.i + εijk ωj(xk
+ − ak) . (6.29a)
We return now to the more general considerations of superposed rigid body motions.
>From (6.25) and (6.11), it follows that
v+ = a.
+ Ω Q x + Q v
= a.
+ Ω Q[QT(x+ − a)] + Q v
= a.
+ Ω (x+ − a) + Q v , (6.30)
or in component form
vi+ = a
.i + Ωij[xj
+ − aj] + Qijvj . (6.30a)
Then,
- 32 -
∂xm+
∂vi+
hhhh = Ωij ∂xm+
∂xj+
hhhh + Qij ∂xn
∂vjhhhh∂xm
+
∂xnhhhh
= Ωij δjm + Qij ∂xn
∂vjhhhh∂xm
+∂hhhh[Qln(xl
+ − al(t))]
= Ωim + Qij ∂xn
∂vjhhhhQln ∂xm+
∂xl+
hhhh
= Ωim + Qij ∂xn
∂vjhhhhQln δlm
= Ωim + Qij ∂xn
∂vjhhhhQmn
= Ωim + Qij Qmn ∂xn
∂vjhhhh , (6.31a)
or in direct notation
∂x+∂v+hhhh = Ω + Q
∂x∂vhhhQT . (6.31)
It follows from (6.31) and the results of section 5 that
D+ = Q D QT ,
W+ = Q W QT + Ω , (6.32)
or, in indicial notation,
dij+ = Qim Qjn dmn ,
wij+ = Qim Qjn wmn + Ωij . (6.32a)
In particular, suppose that the body in its superposed motion is at time t passing through the
configuration κ with angular velocity Ω, so that
Q = I , Q.
= Ω Q = Ω . (6.33)
- 33 -
Then (6.32)
D+ = D , W+ = W + Ω . (6.34)
- 34 -
7. Infinitesimal deformation and infinitesimal strain measures.
We recall from section 1 that a motion of the body is defined by
x = χ(X, t) or xi = χi(XA, t) . (7.1)
The deformation function χ in (7.1) may be written in terms of the relative displacement u (see
section 4), i.e.,
χ = X + u(X, t) or χi = δiA XA + ui(XA, t) . (7.2)
We further recall the relative strain measures
E =21hh(C − I) or EAB =
21hh(CAB − δAB) , (7.3)
and
e =21hh(I − c) or eij =
21hh(δij − cij) , (7.4)
where
C = FT F or CAB = FiA FiB (7.5)
and
c = (F−1)TF−1 = (F FT)−1 = B−1 or cij = FAi−1
FAj−1
. (7.6)
Also, the relative displacement gradient is defined by
H = F − I or ui,A = FiA − δiA . (7.7)
In order to obtain infinitesimal kinematical results from those of the finite theory discussed
in sections 1-5, we introduce a measure of smallnessiiiiiiii by a nonnegative function
ε = ε(t) =K
maxX ∈ R o
sup || u,K || (K = 1, 2, 3) , (7.8)
where sup stands for the supremum (or the least upper bound) of a nonempty bounded set of real
numbers. If f(u,K) is any scalar-, vector-, or tensor-valued function of u,K = u,1, u,2, u,3
defined in a neighborhood of u,K = 0 (K = 1, 2, 3) and satisfying the condition that there exists a
- 35 -
nonnegative real constant C such that || f(u,K) || < Cεn as ε → 0, then we write f = 0(εn) as ε → 0.
Thus, in particular, the components of u,K referred to either ei or EA are of 0(ε) as ε → 0, i.e.,
ui,A = 0(ε) and uB,A = 0(ε) as ε → 0 . (7.9)
The components EAB of the relative strain measure E in (7.3) can also be expressed in
terms of relative displacement gradients (7.9)2 in the form
EAB =21hh(uA,B + uB,A + uM,AuM,B) . (7.10)
Clearly, if terms of 0(ε2) as ε → 0 can be neglected, we can write (7.10) approximately as
EAB =21hh(uA,B + uB,A) = 0(ε) as ε → 0 . (7.11)
Next, consider the expression
FiA(δAj −∂XB
∂uAhhhhhδBj) (7.12)
and substitute for FiA = δiA +∂XA
∂uihhhhh from (7.7)2. Thus,
Expression (7.12) = (δiA +∂XA
∂uihhhhh)(δAj −∂XB
∂uAhhhhhδBj)
= δij +∂XA
∂uihhhhhδAj −∂XB
∂uihhhhhδBj + 0(ε2)
= δij + 0(ε2)
= δij , (7.13)
where in writing (7.13)4, terms of 0(ε2) as ε → 0 have been neglected. Hence, to the order ε2 as
ε → 0, we can identify the coefficient of FiA in (7.12) as the inverse of FiA, i.e.,
FAi−1 =
∂xiBi
∂XAhhhhh . (7.14)
- 36 -
With the use of (7.14) and the chain rule of differentiation, it can be readily verified that
∂xi
∂uAhhhh =∂XB
∂uAhhhhh∂xi
∂XBhhhhh∂XC
∂uBhhhhhδCi)
=∂XB
∂uAhhhhhδBi + 0(ε2) as ε → 0 (7.15)
and similarly
∂xj
∂uihhhh =∂XA
∂uihhhhh∂xj
∂XAhhhhh∂XB
∂uAhhhhhδBj)
=∂XA
∂uihhhhhδAj + 0(ε2) as ε → 0 . (7.16)
It follows from (7.15) and (7.16) that to 0(ε2), it is immaterial whether the partial derivatives of
the displacement field u is taken with respect to xi or XA so that∂x1
∂u1hhhh =∂X1
∂u1hhhh to 0(ε2) as ε → 0.
Hence in dealing with infinitesimal kinematics, it is not necessary to distinguish between
Eulerian and Lagrangian form of kinematical quantities.
>From (7.6)2 and (7.14), we have
cij = FAi−1
FAj−1 = (δAi −
∂XB
∂uAhhhhhδBi)(δAj −∂XC
∂uAhhhhhδCj)
= δij − (∂XB
∂uAhhhhhδBiδAj +∂XC
∂uAhhhhhδCj δAi) + 0(ε2) as ε → 0 . (7.17)
Hence, if terms of 0(ε2) as ε → 0 can be neglected, then
eij =21hh(δij − cij)
=21hh(
∂XB
∂uAhhhhh +∂XA
∂uBhhhhh)δiAδjBε → 0 , (7.18)
so that any differences between the two strain measures disappear upon linearization.
- 37 -
In view of the remark made following (7.15), to the order of ε2 we may display the com-
ponents of the relative displacement gradients either as uA,B or ui,j. For example, if we make the
latter choice, then we may write
ui,j = eij + ωij = 0(ε) ,
eij = 0(ε) , ωij = 0(ε) as ε → 0 , (7.19)
where eij and ωij are defined by
eij =21hh(ui,j + uj,i) , ωij =
21hh(ui,j − uj,i) . (7.20)
We observe here that ui,i = eii = 0(ε) is called the cubical dilatation and that ωii = 0.
We consider now the linearized version of the Cauchy-Green measure CAB, the stretch UAB
and the rotation RiA. The Cauchy-Green measure CAB defined by (7.5)2 is related to EAB by
CAB = δAB + 2EAB. But, when the approximation (7.11) is adopted, then CAB can be written as
CAB = UAC UCB = δAB + 2EAB
= δAB + 0(ε) as ε → 0 (7.21)
and from an examination of
(δAC + EAC)(δCB + ECB) = δAB + 2EAB + 0(ε2) as ε → 0 (7.22)
to the order ε2 we can identify UAB as
UAB = δAB + EAB = δAB + 0(ε) as ε → 0 . (7.23)
Also, by considering an expression of the form
UAB(δBCneglected,then.EQI(7.24)UAB−1 = δAB − EAB = δAB + 0(ε) as ε → 0 .
We now turn to the rotation tensor RiA. Recalling the polar decomposition theorem, the
components of the rotation tensor R can be written as
RiA = FiB UBA−1
. (7.25)
- 38 -
It then follows from (7.7), (7.24) and (7.11) that
RiA = (δiB + ui,B)[δBA −21hh(uA,B + uB,A)]
= δiA +21hh[ui,A − uA,BδiB]
= δiA +21hh(
∂XA
∂uihhhhh −∂xi
∂uAhhhh)
= δiA +21hh(
∂xA
∂uihhhh −∂xi
∂uAhhhh) , (7.26)
where terms of 0(ε2) as ε → 0 have been neglected and where use has been made of the result
(7.15). To the order of approximation considered, we may also write (7.26) as
R = I + Ω or Rij = δij + ωij , (7.27)
where
Ω = −ΩT =21hh(H − HT)j,i) , (7.28)
which is consistent with (7.20)2.
In the remainder of this section, we employ the linearization procedure discussed above and
obtain the linearized version of such expressions as
J = det F = det(xi,A) (7.29)
and its inverse. For this purpose, we first recall that J can be represented as
J =61hh εijk εLMN FiL FjM FkN . (7.30)
Substituting the components of the displacement gradient in terms of ui,A given by (7.7)2 in
(7.30) we get
- 39 -
J =61hhεijk εLMN(δiL + ui,L)(δjM + uj,M)(δk,N + ukN)
=61hhεijk εLMN[δiL δjM δkN + 3δiL δjM uk,N + 0(ε2)]
= 1 + δkN uk,N = 1 + uk,k = 1 + ekk as ε → 0 , (7.31)
where in obtaining (7.31)3 use has been made of the identities61hh = εijk εijk and εijk εijm = 2δkm
and where terms of 0(ε2) as ε → 0 have been neglected. The inverse of (7.31) is given by
J−1 = [1 + eii + . . . ]−1
= 1 − eii + (eii)2 + . . .
= 1 − eii as ε → 0 (7.32)
and again in writing (7.32) terms of 0(ε2) as ε → 0 have been neglected.
(For a discussion of the interpretation of the infinitesimal strain measures, see Supplement
to Part I, Section 7.)
- 40 -
8. The transport theorem.
Let S be an arbitrary part (or subset) of the body B and suppose that S occupies a region Po,
with a closed boundary surface ∂Po, in a fixed reference configuration. Similarly let P, with
closed boundary surface ∂P, be the region occupied by S in the configuration at time t.
Let φ be any scalar-valued or tensor-valued field with the following representations:
φ = φ(x, t) = φ(χ(X, t), t) (8.1)
and consider the volume integral
I =P∫ φ(x, t)dv =
Po
∫ φ(X, t)J dV , (8.2)
where we have used dv = J dV, J = det F > 0. Often we shall encounter an expression of the type
(8.2) and we need to calculate its time derivative dI/dt. Thus, we write
dtdIhhh =
dtdhhh
Po
∫ φ(X, t)J dV
=Po
∫ dtdhhh(φ(X, t)J)dV
=Po
∫ [∂t∂φhhh(X, t)J + φ(X, t)J
.]dV
=Po
∫ [∂t∂φhhh(X, t)J + Jvi,iφ(X, t)]dV
=Po
∫ [∂t
∂φ(X, t)hhhhhhh + φ(X, t)vi,i]JdV
=Po
∫ [φ.
+ φ(X, t)div v]JdV
=P∫[φ
.+ φ(x, t)div (8.3)
- 41 -
where in the fourth of (8.3) we have also used J.
= J vi,i and where
φ.
=∂t
∂φ(X, t)hhhhhhh =∂t
∂φ(x, t)hhhhhhh +∂xi
∂φ(x, t)hhhhhhhvi . (8.4)
It then follows that the time derivative of (8.2)1 is given by
dtdhhh
P∫ φ(x, t)dv =
P∫ [φ
.+ φ(x, t)div v]dv . (8.5)
>From the result (8.5) follows the various expressions given below:
dtdhhh
P∫ φ(x, t)dv =
P∫ [
∂t∂φ(x, t)hhhhhhh +
∂xi
∂φ(x, t)hhhhhhhvi
=P∫ [
∂t∂φ(x, t)hhhhhhh +
∂xi
∂(φ(x, t)vi)hhhhhhhhhh]dv
=P∫ ∂t
∂φ(x, t)hhhhhhhdv +∂P∫ φ(x, t)vinida
=P∫ ∂t
∂φ(x, t)hhhhhhhdv +∂P∫ φ(x, t)v . n da , (8.6)
where the divergence theorem has been used. Next, apply (8.6)4 to a fixediiii spatial region Ph
in the
configuration at time t and write
dtdhhh
Ph∫ φ(x, t)dv =
Ph∫ ∂t
∂φ(x, t)hhhhhhhdv +∂Ph∫ φ(x, t)v . n da
=∂t∂hhh
Ph∫ φ(x, t)dv +
∂Ph∫ φ(x,t)v . n da . (8.7)
The form (8.7)2 is another statement of the transport theorem. It is used when it is convenient to
focus attention on a fixed region of space Ph
at time t and consider the motion of the body over
this region. This form (8.7)2 can also be deduced directly from (8.6)4 without the introduction of
the fixed spatial region Ph
by the following line of argument: Since in the calculation of the first
- 42 -
integral in (8.6)4 the variable x (and hence the spatial region P) is fixed, the operator ∂/∂t com-
mutes with the volume integral in (8.6)4 over a region P and the form (8.7)2 with Ph
replaced by P
follows from (8.6)4. The commuting of ∂/∂t with the volume integral in (8.6)4 may be contrasted
with a similar calculation involving the volume integral in (8.3)2. In the latter integral, since X
(and hence the material region Po, is fixed, the operator d/dt commutes with the volume integral
in (8.3)2 over the region Po.
- 41 -
Part II: Conservation Laws and Some Related Results
- 42 -
1. Conservation of mass.
We recall from section 1 of Part I that the mass of any part P of the body, i.e., M is a con-
tinuous function of its volume and that there exists a scalar mass density ρ such that
M(St) =P∫ ρ dv , P ⊆ R , (1)
where ρ = ρ(x, t) depends on the particular configuration occupying the region of space R and dv
is the element of volume in the present configuration. A statement of the conservation of mass
for any material part of the body in the present configuration is as follows:
dtdhhh
P∫ ρ dv = 0 . (2)
By the transport theorem, (2) can be written asP∫ [ρ
.+ ρvk,k]dv = 0. The last result must hold for
alliii arbitrary parts P; hence, by the argument outlined previously, assuming that ρ is continuously
differentiable we have
ρ.
+ ρvk,k = 0 , (3)
where a superposed dot denotes material derivative, i.e., ρ.
=∂t∂ρhhh + vk ∂xk
∂ρhhhh . The result (3) can
also be expressed as
∂t∂ρhhh + (ρvk),k = 0 or
∂t∂ρhhh + div(ρv) = 0 . (4)
Equation (3) or (4) represents the local equation for conservation of mass. It is also referred to
as spatialiiiiii form of the "continuity equation."
Another form of the principle of conservation of mass may be stated as:
P∫ ρ dv =
Po
∫ ρo dV , (5)
where Po is the material part in the initial reference configuration, dV is the element of volume in
the reference configuration and ρo denotes the mass density in the reference configuration.
Using dv = J dV, J = det F = det xi,A, from (5) we get
- 43 -
ρ = J−1 ρo or ρ = (det F)−1 ρo , ρo = Jρ , (6)
which is the materialiiiiiii form of continuity equation.
- 44 -
Appendix to Sec. 1 of Part II
Theorem: If φ(x, t) is continuous in R and
P∫ φ dv = 0 (1)
for every part P ⊆ R, then the necessary and sufficient condition for the validity of (1) is that
φ = 0 in R . (2)
Proof: Let us first recall the definition of continuity.
Definition: A function φ(x, t) is said to be continuous in a region R if for every x ∈ R and every
ε > 0 there exists a δ > 0 such that | φ(x, t) − φ(ox, t) | < ε whenever | x − ox | < δ.
Sufficiency: If φ = 0 in R, then (1) is trivially satisfied.
Necessity: (Proof by contradiction). Suppose that there is a point ox ∈ P (⊆ R) at which
φo = φ(ox, t)>0. Then, by the continuity of φ, there exists a δ > 0 such that
| φ(x, t) − φo | <2
φohhh for all | x − ox | < δ . (3)
Now let Pδ be the region | x − ox | < δ and Vδ be the volume of this region so that
Vδ =Pδ
∫ dv > 0 . (4)
>From (3) and the fact that φo > 0 we have
φ >2
φohhh in Pδ . (5)
Now let P be a part which contains Pδ, then
Pδ
∫ φ dv >Pδ
∫ 1⁄2 φo dv = 1⁄2 φo Vδ > 0 . (6)
Since the result (6) contradicts (1), it follows that there can exist no point ox such that φ(ox, t) >
- 45 -
0. Similarly, we realize that if φo = φ(ox, t) < 0, then by the continuity of φ there exists a δ > 0
such that
| φ(x, t) − φo | < −2
φohhh for all | x − ox | < δ . (7)
Thus we can conclude that
φ(x, t) <2
φohhh for all x ∈ Pδ . (8)
Hence
Pδ
∫ φ dv <Pδ
∫ 1⁄2 φo dv = 1⁄2 φo Vδ < 0 . (9)
Since (9) contradicts (1), again it follows that no point ox exists for which φ(ox, t) < 0. Combin-
ing this result with that above we conclude that (2) is also a necessary condition for the validity
of (1).
- 46 -
2. Forces and couples. Euler’s laws.
We admit two types of forces, namely the body force per unit mass b and the surface force
(or contact force) per unit area) as follows:
b = b(X,t) = body force per unit mass,
(1)
t = t(X,t ; n) = contact force per unit area.
It is important to note that the surface force t depends on the orientation of the surface area (with
outward unit normal n) upon which it acts. Similar to the definitions (1) above, we could also
admit body couple per unit mass c and surface couple per unit area m, but these are ruled out in
the construction of classical continuum mechanics.
We also note here the following:
momentum per unit mass = v = x.
(2)
moment of momentum per unit mass = x × v = x × x.
We consider now the basic balance laws (also called conservation laws) for momentum and
moment of momentum (also called angular momentum) in the context of purely mechanical
theory. These balance laws, which are known as Euler’s laws, may be stated (in words) as fol-
lows:
IKLfor any part of the body
Rate of change of momentum MNO
IKLon the part
All external forces acting MNO
,
(3)
IKLfor any part of the body
Rate of change of moment of momentum MNO
IKLacting on the part
Moment of all external forces MNO
.
The above laws can be translated into mathematical forms both with respect to the current
- 47 -
configuration (Eulerian form) or with respect to the reference configuration (Lagrangian form).
In this section we limit the discussion to the Eulerian (or spatial) form of the balance laws (3).
Thus, remembering the definitions (2)1,2, the momentum for any part P of the body occupying a
region P with boundary surface ∂P and the moment of momentum for any part P are:
P
∫vdm ,
P
∫x × vdm , (4)
where dm = ρdv is the element of mass. Also, the total force acting on P and the moment of total
force acting on P are:
P
∫bdm +
∂P
∫ tda ,
P
∫(x × b)dm +
∂P
∫ (x × t)da . (5)
Keeping the expressions (4) and (5) in mind, the spatial (or Eulerian) form of the balance
laws (3) are:
dtdhhh
P
∫vdm =
P
∫bdm +
∂P
∫ tda , (6)
dtdhhh
P
∫x × vdm =
P
∫x × bdm +
∂P
∫ x × tda . (7)
In the next few sections of Part II we exploit the implications of (6) - (7) and also derive their
local forms.
- 48 -
3. Further consideration of the stress vector. Existence of the stress tensor and its rela-tionship to the stress vector.
Consider an arbitrary part of the material region of the body B which occupies a part P in
the present configuration at time t. Let P be divided into two regions P1, P2 separated by a sur-
face σ (see Fig. 3.1). Further, let ∂P1, ∂P2 refer to the boundaries of P1, P2, respectively; and let
∂P′, ∂P′′ be the portions of the boundaries of P1, P2 such that
∂P′ = ∂P1 ∩ ∂P ,
∂P′′ = ∂P2 ∩ ∂P . (1)
Thus, a summary of the above description is as follows:
P = P1 ∪ P2 , ∂P = ∂P′ ∪ ∂P′′ ,
∂P1 = ∂P′ ∪ σ , ∂P2 = ∂P′′ ∪ σ . (2)
Now recall the linear momentum principle, i.e.,
dtdhhh
P
∫ ρ x.
dv =P
∫ ρ b dv +∂P
∫ t(n)da (3)
or with the use of dm = ρ dv in the form:
dtdhhh
P
∫ x.
dm =P
∫ b dm +∂P
∫ t(n) da (4)
which holds for an arbitrary material region P ⊆ R. Application of (4) separately to the parts P1,
P2 and again to P1 ∪ P2 = P yields
dtdhhh
P1
∫ x.
dm −P1
∫ b dm −∂P1
∫ t(n)da = 0 , (5)
dtdhhh
P2
∫ x.
dm −P2
∫ b dm −∂P2
∫ t(n)da = 0 (6)
and
dtdhhh
P1∪P2
∫ x.
dm −P1∪P2
∫ b dm −∂P′∪∂P′′
∫ t(n)da = 0 . (7)
- 49 -
The stress vector t(n) in (5) acting over the boundary ∂P1 results from contact forces exerted
byii the material on one side of the boundary (exterior to P1) onii the material of the other side.
Similar remarks hold for the stress vector in (6) and in (7). We emphasize that the stress vector
in (5) over ∂P′ ∪ σ represents the contact force exerted on P1 across the surface, etc. The
appropriate normals associated with t(n) over the surface σ are equal and opposite in sign. To
elaborate, let n be the outward unit normal at a point on σ when σ is a portion of ∂P1. Then, the
outward unit normal at the same point on σ when σ is a portion of ∂P2 is −n.
>From combination of (5) and (6) and after subtraction from (7), we obtain the following
equation:
σ∫ [t(n) + t(−n)]da = 0 (8)
over the arbitrary surface σ. Assuming that the stress vector is a continuous function of position
and n, it follows that
t(n) = −t(−n) or t(x , t ; n) = − t(x , t ; −n) . (9)
According to the result (9), the stress vectors acting on opposite sides of the same surface at
a given point are equal in magnitude and opposite in direction. The result (9) is known as
Cauchy’s lemma.
Again we suppose that a body B is mapped into the present configuration κ, at time t, which
occupies the region R. Consider some interior particle Xo of B having position vector ox in R.
Construct at ox a tetrahedron T, lying entirely within R, and in such a way that the side i is per-
pendicular to the ei-direction (see Fig. 3.2) and inclined plane - with outward unit normal on -
falls in the octant where x1 , x2, x3 are all positive. We refer to the side i of T by Si and to the
inclined plane by S, respectively. Let h denote the height of the tetrahedron, i.e., the perpendicu-
lar distance from ox to the inclined face S, and let S denote the area of this face. Then, the areas
of the three orthogonal faces Si are
Si = S on . ei = S oni (10)
- 50 -
and the volume of the tetrahedron is
Vtet =31hh h S . (11)
Recalling that dm = ρdv, by virtue of the transport theorem and the conservation of mass
(see Sec. 1 of part II), the principle of linear momentum (4) reduces to
P
∫ x..
dm =P
∫ b dm +∂P
∫ t(n)da . (12)
Next, apply (12) to the material region T and obtain
T
∫ x..
dm =T
∫ b dm +∂T
∫ t(n)da . (13)
Observing that the surface integral in (13) represents contributions from all four boundary planes
of T, we have
∂T
∫ t(n)da =i=1Σ3
Si
∫ t(−ei)da +S
∫ t(on)da . (14)
Now according to Cauchy’s lemma in (9)
t(−ei) = − t(ei) (15)
and then with the use of (14) and (15), the linear momentum equation (13) can be rewritten in
the form
T
∫ (x..
− b)dm =S
∫ t(on)da −
i=1Σ3
Si
∫ t(ei)da . (16)
Now recall that dm = ρdv and that ρ is already assumed to be bounded (see Sec. 1 of Part I).
Further, we assume that the fields x..
and b are bounded. Then, since (by a theorem of analysis)
| ∫ f dv | ≤ ∫ | f | dv
where | f | denotes the absolute value of f, we obtain the following estimate for the integral on
the left-hand side of (16):
- 51 -
|
T
∫ ρ(x..
− b)dv | ≤T
∫ | ρ(x..
− b) | dv
=T
∫ K dv
= K*
T
∫ dv = K*
3Shhhh
where use has been made of the mean value theorem for integrals, we have set
K(x ,t) = | ρ(x..
− b) | and K* stands for some specific interior value of K in T. Hence, we may
conclude that there exists a fixed, positive real number K* such that
cT
∫ ρ(x..
− b)dv c ≤ K*
3Shhhh . (17)
Next, assume that the stress vector field is continuous in both x and n. Then, by the mean value
theorem for integrals, the two surface integrals in (16) yield:
i=1Σ3
∫Si
t(ei)da = ti*Si = ti
* S oni (18)
and
S
∫ t(on)da = t(on)* S , (19)
where in writing (18)2 use has been made of (10) and where ti* and t* stand for some specific
interior values (at the point ox) of the stress vectors on the respective faces Si and on the plane S
of T. Each stress vector ti* acts on that side of coordinate plane i which is associated with the
outward unit normal ei. It follows from (16), (17), (18) and (19) that (since K* is fixed)
31hh K*Sh ≥ |
T
∫ ρ(x..
(on)da −
i=1Σ3
∫Si
t(ei)da |
= | t(on)* S − ti
* S oni |
= S | t(on)* − ti
*oni | .
- 52 -
Therefore,
| t(on)* − ti
*oni | ≤
31hh K*h . (20)
Consider now a sequence of tetrahedra T1, T2, ..., of diminishing heights h1 > h2 > ..., each
member of which is similar to T with three mutually orthogonal faces having outward unit nor-
mals −ei and an inclined plane with outward unit normal on. Next, apply (20) to each member of
the sequence and, in the limit as h→ 0, obtain
| t(on)* − ti
*oni | ≤ 0 , (21)
where the stress vectors in (21) are evaluated at the point ox which is the common vertex of the
family of tetrahedra. It follows from (21) that
t(on)* = ti
*oni . (22)
Since (22) must hold at any point ox and corresponding to any direction on, without ambiguity,
we suppress henceforth the designation star, replace ox by x, on by n and write
t(n) = tini . (23)
We now define tki by
tki = ti. ek or ti = tki ek . (24)
Let tk denote the components of the stress vector t(n), i.e., tk = t(n). ek. Then, from (23) and (24)
we have
tk = t . ek = tini. ek = tkini , (25)
where in (25)1 we have written for simplicity t in place of t(n). Thus, if t = t(x , t ; n) is the stress
vector at the point x acting on a surface whose outward unit normal is n, then it is clear that
tki = tki(x , t) are defined at x and are independent of n. The relation (25) establishes the
existenceiiiiiiii of the set of quantities tki. It remains to show that tki are Cartesian components of a
second order tensor. To this end, consider the transformation of two sets of Cartesian coordi-
- 53 -
nates and recall
xk′ = ajkxj , ek′ = ajkej , nk′ = ajknj , etc. , (26)
where
aik = ei. ek′ . (27)
From
t = tkek = tk′ek′ = tkiniek = tki′ni′ek′ (28)
and with the use of ni′ = arinr , ek′ = ask es where ari are components of an orthogonal tensor, we
obtain (tsr − ariasktki′)nres . But es are linearly independent. Therefore,
(tsr − ariasktki′)nr = 0 , (29)
which holds for all n and the quantity in parentheses is independent of n. Hence,
tsr = askaritki′ , (30)
which establishes the tensor character of tki. The second order tensor whose Cartesian com-
ponents are defined by (24) is called the Cauchy stress tensor.
With reference to a rectangular Cartesian system of spatial coordinates, consider a paral-
lelepiped shown in Fig. 3.3 and recall the formula
ti = tkiek (31)
In (31), ti is the stress vector acting on the face whose outward unit normal is ei. For example,
on the front face of the parallelepiped in Fig. 3.3, the outward unit normal coincides with e1; and
thus, from (31) with i = 1, the stress vector on this face is given by
t1 = tk1ek = t11e1 + t21e2 + t31e3 .
Keeping the above in mind, an examination of the subscripts of the stress tensor tki in (31) easily
reveals that (i) the second index "i" refers to the stress vector ti on the face whose outward unit
normal is ei, while (ii) the first index "k" refers to the component of ti in the coordinate
- 54 -
directions. Utilizing this convention, which stems from the definition of (24), the tractions (i.e.,
the forces per unit area) are sketched in Fig. 3.3.
- 55 -
4. Derivation of spatial (or Eulerian) form of the equations of motion.
We derive in this section the spatial (or Eulerian) form of the equations of motion from the
integral balance laws (6) and (7) of section 2. We begin with the balance of linear momentum
(6) in section 2 of Part II, which for the present purpose can be written as (recall that dm = ρdv):
dtdhhh
P
∫ρvdv =
P
∫ρbdv +
∂P
∫ t(n)da . (1)
By the transport theorem discussed in Part I, the left-hand side of (1) is given by
LHS of (1) =
P
∫[(ρv)hhhh.
+ ρvdivv] dv
=
P
∫[ρ.v + ρv
.+ ρvdivv] dv
=
P
∫ρv.
+ v[ρ.
+ ρdivv] dv (2)
=
P
∫ρadv ,
where in arriving at the results, (2)4, we have also used the local conservation of mass given by
(3) of section 1. Next, we consider the surface integral in (1), substitute the relation (23) of sec-
tion 3, namely t = tini, and use the divergence theorem obtain
∂P
∫ tda =
∂P
∫ tinida =
P
∫ti,i dv , (3)
where a comma following a subscript indicates partial differentiation.
After substitution of the results (2) and (3) into (1), the resulting equation can be put in the
form
P
∫[ti,i + ρb − ρa] dv = 0 , (4)
- 56 -
which must hold for all arbitrary material volumes. Then, since the integrand is a continuous
function, by the usual procedure we arrive at the local form
ti,i + ρb = ρa . (5)
Next, we turn to the balance of moment of momentum (7) in section 2 of Part II, which can
be rewritten as
dtdhhh
P
∫(x × ρv) dv =
P
∫(x × ρb) dv +
∂P
∫ x × t(n) da (6)
Again by the transport theorem, the left-hand side of (6) can be shown to yield
LHS of (6) =
P
∫x × ρa dv . (7)
Similarly, the surface integral in (6) after substitution from (23) of section 3 and the use of the
divergence theorem gives
∂P
∫ x × t(n) da =
∂P
∫ (x × tini) da =
P
∫(x × ti),i dv
(8)
=
P
∫[(x,i × ti) + x × ti,i] dv
Introduction of (7) and (8) into (6) results in
P
∫[(x,i × ti) + x × (ti,i + ρb − ρa)] dv = 0 ,
or
P
∫x,i × ti dv = 0 , (9)
- 57 -
where in obtaining the last result we have also used (5). Since (9) must hold for all arbitrary
material volumes P and since the integrand is continuous, by the usual argument we conclude
that
x,i × ti = 0 ,
or equivalently
ei × ti = 0 . (10)
The two results (5) and (10) represent the consequences of the balance of linear momentum
and moment of momentum. Although the foregoing derivation in vectorial form leading to
equations of motion (5) and the restriction (10) is simple and attractive, it conceals some of the
features of the stress tensor introduced earlier in section 3. Thus, we now recall the relation (24)
for ti = tkiek and substitute this into (5) and (10) to obtain the alternative forms of the equations
of motion and the restriction on ti. With the use of the identities
ti,i = (tkiek),i = tki,iek ,
(11)
ei × ti = ei × tkiek = εikjtkiej ,
the component forms of (6) and (10) result in
tki,i + ρbk = ρak , (12)
tki = tik . (13)
Thus, it is easily seen that the vector equation (5) is equivalent to the three scalar equations of
motion (12). Similarly, the restriction (10) on the three vectors ti implies the symmetry of the
stress tensor tki. This last observation reveals the fact that the three scalar equations of motion
involve only six components of the stress instead of nine.
- 58 -
5. Derivation of the equations of motion in referential (or Lagrangian) form.
In previous developments, the stress vector t (and therefore the stress tensor tij) acting on P
are measured per unit area of surfaces in the present configuration at time t. In view of the
transformation x = χ (X , t) all surfaces in R can be mapped into corresponding surfaces in Ro
occupied by the body in the reference configuration. For some purposes, it is more convenient to
measure the stress vector and the stress tensor acting on P per unit area of surfaces in Po.
Corresponding to an arbitrary part P with boundary ∂P in the present configuration, we have a
part Po with boundary ∂Po in the reference configuration (see Fig. 1.1). We denote the outward
unit normal to ∂Po by N, where
N = NAeA . (1)
We denote the stress vector acting on ∂P, but measured per unit area of the surface ∂Po in the
reference configuration by p.
In terms of quantities measured in the reference configuration, the momentum principles
(i.e., linear and moment of momentum) are:
Po
∫ ρo a dV =Po
∫ ρo b dV +∂Po
∫ p dA , (2)
and
Po
∫ (x × ρo a)dV =Po
∫ (x × ρo b)dV +∂Po
∫ x × p dA (3)
where p = p(X , t ; N) depends on position, time and the unit normal N to ∂Po. By a procedure
similar to that used previously (Part II: Sec. 3), we can prove that
p = NApA , pA = piAei , (4)
p = piA NA ei , (5)
p = P N . (5a)
The vectors pA (A = 1,2,3) represent stress vectors acting across surfaces at a point P in the
- 59 -
present configuration κ which were originally coordinate planes XA = const. through the
corresponding point Po in the reference configuration and measured per unit area of these planes
(Fig. 5.1). Also, the components piA (i.e., of the stress tensor P) represent surface forces acting
in the present configuration, but measured per unit area of an XA-plane in the reference
configuration and resolved parallel to the ei-directions. Introducing (5) into (2), by a procedure
similar to that used previously (Part II, Sec. 4), we obtain
piA,A + ρobi = ρoai or pA,A + ρob = ρoa , (6)
Then, from (3), (5), and after using (6), we also obtain
piAxj,A = pjAxi,A . (7)
Introducing the notation sAB through
piB = xi,AsAB , (8)
it follows from (7) that
sAB = sBA . (9)
The last two are called, respectively, the symmetric and the nonsymmetric Piola-Kirchhoff
stress.
Also, it can be verified that
JT = P FT = F S FT or Jtik = xi,ApkA = xk,ApiA = xi,Axk,BsAB , (10)
which relates the Cauchy stress T to the stress tensors P and S. To verify the truth of (10), we
may start with
df = t da = p dA (11)
which represents an element of contact force acting on the current configuration of the body B in
terms of both the spatial and referential description of the stress vector. Using the derived rela-
tion between the area elements da and dA, namely n da = J F−T N dA, we have
- 60 -
df = t da = T n da
(12)
= p dA = P N dA .
Hence (J T F−T − P)N dA = 0 for all N and we may conclude that J T F−T − P = 0 or
J T = P FT . (13)
Recalling (8) or P = F S, we are led to the results (10)1.
The coordinate-free forms of the various results between (6)-(10) may be displayed as fol-
lows:
Div P + ρo b = ρo v.
, (6a)
P FT = F PT , (7a)
P = F S , (8a)
S = ST . (9a)
- 61 -
6. Invariance under superposed rigid body motions.
It was established previously that a particle X of a body, which in the reference
configuration is at X and at time t in κ occupies the place x = χ(X, t), under superposed rigid
motions of the whole body at a different time t+ = t + a occupies in κ+ the place x+ = χh
+(X, t+)
specified by
x+ = a + Q x or xi+ = ai + Qijxj . (1)
In (1), a is a vector-valued function of t, Q is a proper orthogonal tensor-valued function of t and
a is a constant. The vector a can be interpreted as a rigid body translation and Q as a rotation
tensor.
We have already studied how various kinematical quantities transform under superposed
rigid body motions and shall presently extend these considerations to the dynamical quantities
which appear in the equations of motion, namely
tij,j + ρbi = ρv.i , tij = tji . (2)
However, we first need to establish some further kinematical results:
(a) Further kinematical results:
Recall that an element of area dA of a material surface having outward unit normal vector
N in the reference configuration of a body is deformed at time t into the element of area da
whose outward unit normal vector is n. Then,
dak = J∂xk
∂XKhhhhhdAK , (3)
where
dak = da nk , dAK = dA NK . (4)
Under the motion χh
+ the element of area dA is deformed into da+ and N is deformed into n+ so
that
dak+ = J+
∂xk+
∂XKhhhhhdAK , (5)
- 62 -
and
dak+ = da+nk
+ . (6)
It follows from (1)2 that
∂xk
∂xi+
hhhh = Qij ∂xk
∂xjhhhh = Qijδjk = Qik (7)
and consequently
FiA+ =
∂XA
∂xi+
hhhhh j (8)
i.e.,
F+ = Q F . (9)
Therefore,
J+ = det F+ = det(Q F)
= J .
= +1 det F
= (det Q)(det F)
(10)
Also, by the law of conservation of mass,
ρo = ρJ = ρ+J+ , (11)
which together with (10)5 implies
ρ+ = ρ . (12)
Returning again to (1)2, we see that
Qik xi+ = Qik ai + QikQijxj
= Qikai + xk ,
= Qikai + δkjxj
(13)
- 63 -
xk = Qik(xi+ − ai) or x = QT(x+ − a) . (14)
Therefore
∂xi+
∂xkhhhh = Qik , (15)
and consequently
∂xk+
∂XKhhhhh =∂xj
∂XKhhhhh∂xk
+
∂xjhhhh =∂xj
∂XKhhhhhQkj (16)
which, in direct notation, reads
(F+)−1 = F−1QT (17)
and could also have been deduced by taking the inverse of both sides of (9).
Substituting the results (10)5 and (16)2 into (5) yields
dak+ = J
∂xj
∂XKhhhhhQkjdAK
= Qkjdaj , (18)
where (3) has been used in completing the last step. The combination of (18)2 with (4)1 and (6)
gives
da+nk+ = Qkjda nj . (19)
Now square both sides of (19) and remember that n and n+ are unit vectors and Q is a proper
orthogonal tensor so that
(da+)2nk+ nk
+ = (da+)2
= Qkj nj Qkl nl (da)2
= δjl nj nl (da)2
= nj nj (da)2 = (da)2 . (20)
- 64 -
Since area is always a positive number, it follws from (20)5 that
da+ = da (21)
and hence from (19) that
nk+ = Qkj nj or n+ = Q n . (22)
It is worthwhile noticing that we have been able to deduce the behavior of F, F−1, J, ρ, da
and n under superposed rigid body motions without making any additional physical assumptions
in (a).
(b) The stress vector and the stress tensor.
It is clear from the developments of section 2 that not all kinematical quantities transform
according to formulae of the type
u+ = Q(t)u , U+ = Q(t)U QT(t) (23)
under superposed rigid body motions, where u and U in (23) stand for a vector and a second
order tensor field, respectively. We investigate now the relationships between† t and t+ and
between T and T+. For convenience, we recall the formulae for ti and tij, i.e.,
t = t(X, t; n) , ti = tijnj , (24)
t+ = t+(X, t+; n+) , ti+ (25)
The mechanical fields which enter the linear and moment of momentum principles are the
stress vector and the body force per unit mass. We first consider the former, which in the motion
χ(X, t) is a vector field defined by (24)1. Consider now a second motion which differs from the
first only by superposed rigid body motions specified in the form (1). The second motion
imparts a change in the orientation of the body, so that the outward unit normal vector n to the
same material surface (in the configuration κ) becomes the unit normal vector n+ (in the
configuration κ+). The stress vector in the second motion takes the form indicated by (25)1 andhhhhhhhhhhhhhhh
† Our development follows Naghdi (1972, pp. 484-486). See also Green and Naghdi (1979).
- 65 -
we have already seen that the outward unit normal vector n+ transforms according to (22). The
geometrical properties of the transformation (1) were discussed in section 2. Keeping these in
mind and recalling that t is linear in n, it is reasonable to expect for the stress vector t+(X, t; n+)
(i) to have the same magnitude as t(X, t; n) and (ii) to have the same relative orientation to n+ as
t has relative to n. These remarks lead us to introduce the following assumption:
t+ = Q t , ti+ = Qijtj . (26)
We observe that (26) implies that
| t+ | 2 = t+ . t+ = ti+ ti
+ = Qij Qik tj tk = δjk tj tk = tj tj = | t | 2 (27)
and, recalling (22), we also have
t+ . n+ = ti+ ni
+ = Qij Qik tj nk = δjk tj nk = tj nj = t . n . (28)
Now
t+ . n+ = | t+ | | n+ | cosθ+ = | t+ | cosθ+ (29)
and
t . n = | t | | n | cosθ = | t | cosθ ,
which together with (27)7 and (28)5 lead to
cosθ+ = cosθ . (30)
The results (27) and (30) verify the motivating remarks made prior to (26) to the effect that (i)
the magnitude of t is the same as the magnitude of t+ and (ii) the magnitude of the angle between
t and the outward unit normal n is the same as the magnitude of the angle between t+ and n+.
We now turn attention to the stress tensor. >From the results (24)2, (25)2 and (22) we have
ti+ = tij
+ nj+ = tij
+ Qjk nk . (31)
But, by assumption (26), we also have
ti+ = Qij tj = Qij tjk nk . (32)
- 66 -
Combination of (31) and (32) yields
(tij+ Qjk − Qij tjk)nk = 0 . (33)
The last result must hold for all nk and the coefficient of nk is independent of nk. Therefore
tij+ Qjk − Qij tjk = 0 ,
tij+ Qjk Qmk − Qmk Qij tjk = 0 ,
tim+ = Qij Qmk tjk ,
and hence
tij+ = Qim Qjn tmn (34)
or, in direct notation,
T+ = Q T QT . (35)
A scalar, vector or tensor quantity which under superposed rigid body motions transforms as
(12), (22) or (35), respectively, is called objectiveiiiiiiii. Of course, not all physical quantities are
objective. For example
W+ = Q W QT + Ω
v+ = a.
+ Q.
x + Q x.
,
(36)
are clearly not objective.
(c) Body forces and accelerations.
The equations of motion for the two motions χ and χh+ of the body are, respectively, given
by
∂xj
∂tijhhhh = ρ(v.i − bi) ,
(37)
∂xj+
∂tij+
hhhh = ρ+(v.
i+ − bi
+) .
- 67 -
But
∂xj+
∂tij+
hhhh =∂xk
∂hhhh (Qim Qjn tmn)∂xj
+
∂xkhhhh
= Qim Qjn ∂xk
∂tmnhhhhh Qjk
= Qim δkn ∂xk
∂tmnhhhhh
= Qim ∂xn
∂tmnhhhhh , (38)
where (15) has been employed in (38)2. It follows from (38)4 and (37) that
ρ+(v.
i+ − bi
+) = Qim ρ(v.
m − bm) , (39)
which with the aid of (12) becomes
v.
i+ − bi
+ = Qim (v.
m − bm) (40)
or
v. + − b+ = Q(v
.− b) .
References
1. Naghdi, P.M. 1972 The theory of shells and plates. In S. Fl..ugge’s Handbuch der Physik,Vol. VIa/2 (edited by C. Truesdell), Springer-Verlag, 425-640.
2. Green, A.E. & Naghdi, P.M. 1979 A note on invariance under superposed rigid bodymotions. J. Elasticity 9, 1-8.
- 68 -
7. The principle of balance of energy.
We begin by assuming the existence of a scalar potential function per unit mass ε = ε(x,t),
called the specific internal energy. The internal energy for each part P in the present
configuration is defined by the volume integral
P
∫ ρ ε dv . (1)
We introduce a scalar field r = r(x,t) per unit mass per unit time, called the specific heat supply
(or heat absorption), as well as the heat flux across a surface ∂P (with the outward unit normal n)
by the scalar h = h(x,t;n) per unit area per unit time. The integrals
H =P
∫ ρ r dv −∂P
∫ h da , (2)
where ∂P is the boundary of P, defines the heat per unit time entering the part P in the present
configuration. The first term on the right-hand side of (2) represents the heat transmitted into P
by radiation and the second term the heat entering P by conduction. We also recall that the
kinetic energy for each part P in the present configuration is defined as
P
∫ 21hh ρ v . v dv . (3)
We record now the rate of work by the body and contact forces, i.e., R = Rb + Rc, where
Rb =P
∫ ρ b . v dv , Rc =∂P
∫ t . v da . (4)
In (4), the scalar b . v which occurs in the volume integral represents the rate of work per unit
mass by the body forces. Similarly, t . v is a scalar representing rate of work per unit area by the
contact force t. Each of the integrals in (4) is a rateiiii ofii workiiiii term and thus has the dimension of
"work per unit time".
With the foregoing background, the law of balance of energy may be stated as follows: The
rate of increase of internal energy plus kinetic energy is equal to the rate of work by body force
and contact force plus energies due to heat per unit time entering the body. Thus, we write
- 69 -
dtdhhh
P
∫ ρ[ε +21hh v . v] dv =
P
∫ ρr dv +P
∫ ρb . v dv
+∂P
∫ t . v da −∂P
∫ h da . (5)
We observe that the negative sign in front of the last integral in (5), as well as in (2), is in accord
with the convention that heat h = h(x,t;n) is assumed to flow into the surface in the direction
opposite to that of the outward unit normal to ∂P.
By application of (5) to an elementary tetrahedron, it can be shown that h(n) = −h(−n) or
h(x,t;n) = −h(x,t;−n), together with
h = qini = q . n , q = qiei . (6)
The last surface integral in (5) can then be transformed into a volume integral as follows (using
the divergence theorem):
∂P
∫ h da =∂P
∫ qini da =P
∫ qi,i dv . (7)
Similarly, using t = tkiniek and the divergence theorem
∂P
∫ t . v da =∂P
∫ (ti. v),i dv
=P
∫ [tki,ivk + tkivk,i] dv . (8)
By the transport theorem, the left-hand side of (5) can be expressed
dtdhhh
P
∫ ρ[ε +21hh v . v] dv =
P
∫ ρ(ε.
+ v . v.) + (ε +
21hh v . v)(ρ
.+ ρvm,m) dv
=P
∫ ρ(ε.
+ v . v.) , (9)
where in obtaining the last result conservation of mass has been used.
- 70 -
Now, substitute (7), (8) and (9) into (5) and rearrange the terms to obtain
0 =P
∫ ρ r − qi,i − ρ ε.
+ tkivk,i + vk[tki,i + ρ(bk − v.k)] dv ,
and after using also the equations of motion we get
P
∫ ρ r − qi,i − ρ ε.
+ tkivk,i dv = 0 , (10)
which holds for all arbitrary parts P. Hence, it follows that the local energy equation is
ρ r − qi,i − ρ ε.
+ P = 0 , (11)
where the stress power P is defined by
P = tkivk,i = tkidki , (12)
and where we have made use of the fact that vk,i = dki + wki and tki is symmetric.
- 71 -
Part III: Examples of Constitutive Equations and Applications
Inviscid and viscous fluid and nonlinear and linear elasticity
- 72 -
Examples of Constitutive Equations
Generally materials are classified as solids or fluids. However, we also encounter materials
which do not fall into either of these categories and which possess properties common to both
solids and fluids. A strict definition of these classifications is not attempted here, but we mention
here some of the main characteristic features of solids and contrast these with corresponding
features in the case of fluids. One of the main characteristic features of a solid is that it has a
reference state (or a reference configuration); and, in general, the deformed state (or
configuration) is not too far from the reference state. Moreover, regardless of the amount of
deformation, to some extent a solid remembers its reference state. A fluid, on the other hand,
does not possess a reference state (or configuration) and does not necessarily remain near such a
state in any motion.
1. Inviscid fluid
In hydrodynamics, an inviscid or an ideal fluid (including an ideal gas) is conceived of as a
medium possessing the following properties: (i) it cannot sustain shearing motion and (ii) the
stress vector (or the force intensity) acting on any surface in the fluid is always directed along
the normal to the surface, i.e., t is parallel to n. However, it is desirable to begin from a more
general viewpoint. Thus, we assume that an inviscid fluid is characterized by a constitutive
equation of the form
T = T(ρ) or tij = tij(ρ) . (1.1)
In (1.1), the response function T is a single-valued function of its argument and ρ is the mass
density of the fluid. Morever, it is instructive to consider an even more general assumption and
suppose the constitutive response function to depend also on the velocity vector†v. This is
because at this stage it is not clear why the explicit dependence on v (or even x) should be ruled
hhhhhhhhhhhhhhh† We rule out inclusion of the velocity gradient which gives rise to a shearing motion. The latter is not compati-
ble with the properties of an ideal fluid stipulated in the opening paragraph of section 1.
- 73 -
out. With this background, we begin by examining the more general constitutive assumption:
T = T(ρ, v) or tij = tij(ρ, vk) . (1.2)
It is convenient to recall here (from Parts I & II) that under a superposed rigid body motion
the position x moves to x+ in accordance with
x+(τ) = a(τ) + Q(τ)x , (τ ≤ t)
(1.3)
v+(τ) = a′(τ) + Q(τ)v(τ) + Q′(τ)x ,
where Q′ = Ω Q, Ω is the rigid body angular velocity, the temporary notation Q′(τ) is defined by
Q′(τ) =dτdhhhQ(τ) with τ being real. Consider now a special case of (1.3) for which at time τ = t,
a′(t) = a.(t) is a constant vector and Ω is a constant Ωo, such that at time t
a.(t) = c , Q
.(t) = Ωo Q(t) ,
where c is a constant vector and Ωo is a constant rigid body angular velocity tensor. We note for
later reference that when the above special motion represents a rigid body translational velocity,
we obtain
Q(t) = I , Q.
(t) = 0 ,
(1.4)
v+ = c + v .
Also, when the special motion involves only a constant rigid body angular velocity, we have
c = 0 , Q(t) = I , Q.
(t) = Ωo ,
(1.5)
v+ = v + Ωox , L+ = L + Ωo
We now examine (1.2)1 under a constant rigid body translational velocity of the form (1.4).
Since the constitutive equation (1.2)1 must hold for all motions, including a s.r.b.m., and since
on physical grounds the response function T (but not its arguments) must remain unaltered under
such superposed rigid body motions, we have
- 74 -
T+ = T(ρ+, v+) = T(ρ+, v + c) . (1.6)
Now under the special motion (1.4), the stress tensor transforms as
T+ = T . (1.7)
Substituting (1.2)1 and (1.6)2 with (1.7) we get
T(ρ, v) = T(ρ, v + c) , (1.8)
which must hold for all constant values c. It follows that (1.8) can be satisfied only if the depen-
dence of T on v is suppressed and we are left with the original assumption (1.1). It should be
evident that the same type of conclusion can be reached if we had initially included in (1.2) also
dependence on the current position x.
Once more we examine the constitutive equation (1.1) under a general s.r.b.m., and recall
that under such motions
T+ = Q T QT .
Introducing the assumption (1.1) into the above invariance requirement, we obtain
T(ρ+) = Q T (ρ) QT , (1.9)
which must hold for all proper orthogonal tensors Q. Since the condition (1.9) is also unaltered
if Q is replaced by − Q, it follows that (1.9) must hold for all orthogonal Q and not just the
proper orthogonal ones. Hence, T(ρ) is an isotropic tensor and can only be a scalar multiple of
the identity tensor I (see the Supplement), i.e.,
T = T(ρ) = −p(ρ)I or tij = tij = −p(ρ)δij . (1.10)
In (1.10), p is a scalar function called the pressure and the minus sign is introduced by conven-
tion, in order to conform to the traditional form in which the stress tensor T = −pI for an inviscid
fluid is displayed in the fluid dynamics literature.
Substitution of the result (1.10) into the relation t = T n (see Part II) yields
t = −p(ρ) n or tk = −p(ρ)nk , (1.11)
- 75 -
which shows that the stress vector acting on a surface in an inviscid fluid is always directed
along the normal to the surface. In most books on fluid dynamics, (1.11)1 is taken as a defining
property of an inviscid fluid, as was also indicated in the opending paragraph of this section. If
(1.11) instead of (1.1) is taken as a starting point, then with the use of t = T n we have
T n = −p(ρ)n or (T + pI)n = 0 which must hold for all outward unit normal n, and we may con-
clude that T = −pI in agreement with the earlier result (1.10) for the stress tensor in an inviscid
fluid.
We are now in a position to obtain the differential equation of motion for the determination
of the velocity field v in an inviscid fluid. Thus, after substituting (1.10) into the equations of
motion (see Part II), we arrive at
−grad p + ρb = ρv.
or p,i + ρbi = ρv.i . (1.12)
For a specified body force b, the vector equation (1.12)1 [or (1.12)2], together with the equation
for mass conservation represent four scalar equations for the determination of the four unknown
(p, v). The simple structure of (1.12) at first sight may be somewhat deceptive: these equations
are difficult to solve by analytical procedures (or even numerical methods). This is because of
the nonlinearity due to the convective terms in v..
- 76 -
2. Viscous fluid.
Although our main objective here is the development of linear constitutive equations for a
(Newtonian) viscous fluid, it is enlightening to start with a more general constitutive assumption.
For a viscous fluid we assume that the stress tensor T depends on the present value of the mass
density ρ, the velocity vector v and the velocity gradient tensor L. Thus, we write
T = Thh
(ρ, v, L) or tij = th
ij(ρ, vk, vk,l) , (2.1)
where Thh
is a single-valued function of its arguments and satisfies any other continuity or dif-
ferentiability conditions that may be required in subsequent analysis. Alternatively, for conveni-
ence and without loss in generality, we recall that L = D + W and rewrite (2.1) as
T = T(ρ, v, D, W) or tij = tij(ρ, vk, dkl, wkl) . (2.2)
The constitutive assumption (2.2)1 [or (2.2)2] must hold for all admissible motions, includ-
ing such superposed rigid body motions as those specified by (1.4) and (1.5). First, by a pro-
cedure parallel to that discussed in the previous section (see the development between Eqs.
(1.6)-(1.8) of section 1) we may suppress the dependence of the response function in (2.2)1 on v
and arrive at
T = T(ρ, D, W) or tij = tij(ρ, dkl, wkl) . (2.3)
Next, from the invariance requirement for T, namely T+ = Q T QT, we obtain the restriction
T(ρ+, D+, W+) = Q T(ρ, D, W) QT , (2.4)
which must hold for all proper orthogonal tensors Q. Consider now the special superposed rigid
body motion specified by (1.5). Under this special motion (2.4) becomes
T(ρ, D , W + Ωo) = T(ρ, D, W) , (2.5)
so that the response function T cannot depend on W and the constitutive assumption (2.3) is
reduced to
T = T(ρ, D) or tij = tij(ρ, dkl) , (2.6)
- 77 -
where T in (2.6) is a different function from that in (2.3). By imposing on (2.6) the invariance
requirement we arrive at
T(ρ+, D+) = Q T(ρ, D)QT . (2.7)
The restriction (2.7) must hold for all proper orthogonal tensors Q. Since the condition (2.7) is
also unaltered if Q is replaced by − Q, it follows that (2.7) holds for all orthogonal Q and not
just the proper orthogonal ones.
A viscous fluid characterized by the nonlinear constitutive equation (2.6)1 is known as the
Reiner-Rivlin fluid. We do not pursue further discussions of the general forms (2.6)1, but in the
remainder of this section consider in some detail the (Newtonian) linear viscous fluid as a special
case in which the response function T is a linear function of D with coefficients which depend on
ρ.
In the discussion of the linear viscous fluid, it is convenient, in what follows, to carry out
the details of the development of the constitutive equations in terms of their tensor components.
Thus, for a linear viscous fluid, we specifiy tij on the right-hand side of (2.6)2 to be linear in dkl
in the form
tij(ρ, dkl) = aij + bijkl dkl , (2.8)
the coefficients aij and bijkl are functions of ρ and satisfy the symmetry conditions
aij = aji , bijkl = bjikl , bijkl = bijlk , (2.9)
where the condition (2.9)3 arises from the symmetry of dkl and is assumed without loss in gen-
erality.
The response function (2.8) must satisfy the invariance condition (2.7) which in component
form is
tij(ρ, dkl+) = Qim Qjn tmn (ρ, dkl) , (2.10)
where in recording the above we have also used ρ+ = ρ in the argument of the left-hand side of
(2.10). Introducing (2.8) into (2.10) we obtain the restriction
- 78 -
aij + bijkl Qkp Qlq dpq = Qim Qjn (amn + bmnpq dpq) (2.11)
which holds for all arbitrary dkl. Considering separately the cases when dkl = 0 and dkl ≠ 0, it
follows from (2.11) that
aij = Qim Qjn amn , (2.12)
and
bijkl = Qim Qjn Qkp Qlq bmnpq ,
where in obtaining (2.12)2 we have made use of the orthogonality property of Qij. Now accord-
ing to the theorems on isotropic tensors (see the Supplement) the coefficients aij and bijkl are,
respectively, isotropic tensors of order two and four and hence must have the forms
aij = −pδij (2.13)
bijkl = λδij δkl + µδik δjl + γδil δjk ,
where p, λ, µ, γ are functions of ρ only and the use of −p instead of p in (2.13)1 is for later con-
venience and in order to conform to known expressions in the fluid dynamics literature. If we
appeal to the symmetry condition (2.9)2, then the right-hand side of (2.13)2 must be symmetric
in the pair of indices (ij) and (2.13)2 reduces to
bijkl = λδij δkl + µ(δik δjl + δil δjk) . (2.14)
It should be noted that the expression (2.14) implies a further symmetry restriction so that bijkl
satisfies
bijkl = bklij , (2.15)
in addition to (2.9)2,3. Substitution of (2.13)1 and (2.14) into (2.8) finally yields
tij = −pδij + λδij dkk + 2µdij , (2.16)
as the constitutive equation for a (Newtonian) linear viscous fluid. The scalar p in (2.16) is the
pressure and λ, µ are called the viscosity coefficients.
- 79 -
The differential equations of motion for a (Newtonian) linear viscous fluid in terms of the
velocity field can be obtained by substituting (2.16) into Cauchy’s stress equations of motion and
leads to
−p,i + λdkk,i + 2µdij,j + ρbi = ρv.i .
After recalling the kinematical relations
dkk = vk,k , dij =21hh(vi,j + vj,i)
and making use of vj,ij = vj,ji, the above differential equations of motion become
−p,i + (λ + µ) vk,ki + µvi,kk + ρbi = ρv.i , (2.17a)
or
−∇p + (λ + µ)∇(∇ . v) + µ∇2v + ρb = ρv.
. (2.17b)
The system of differential equations (2.17) and the mass conservation represent four scalar equa-
tions for the determination of the four unknowns, (p, v). It should be noted in the absence of
viscous affects (λ = µ = 0), (2.17) reduces to those for an inviscid fluid (see Eqs. (1.12)).
- 80 -
3. Elastic solids: Nonlinear constitutive equations.
We consider here general constitutive equations for an elastic solid in the context of the
purely mechanical theory.
A material is defined by a constitutive assumption. Moreover, an elastic material in the
purely mechanical theory may be defined in terms of a constitutive equation for the stress. We
return to this later but first we need to establish some preliminaries.
We recall the expression for the rate of work by the body and contact forces as follows (see
Sec. 7 of Part II):
R = Rb + Rc , (3.1)
Rb =P
∫ ρb . v dv , Rc =∂P
∫ t . v da . (3.2)
Also, the local equations of motion (in terms of the symmetric stress tensor of Cauchy) are:
tki,i + ρbk = ρv.k , tki = tik . (3.3)
Consider now the expression for Rc, use the divergence theorem to transform the surface
integral into a volume integral and make use of the equations of motion (3.3)1 to obtain:
Rc =∂P
∫ t . v da =∂P
∫ tkvkda =∂P
∫ tkinivkda =P
∫(tkivk),idv
=P
∫tki,ivkdv +P
∫tkivk,idv
=P
∫ρv.kvkdv −
P
∫ρbkvkdv +P
∫tkivk,idv . (3.4)
Then, after using (3.2)1
Rc =P
∫ρv.kvkdv − Rb +
P
∫tkivk,idv . (3.5)
From this result and (3.1) follows that
R =dtdhhhK(St) +
P
∫tkivk,idv , (3.6)
- 81 -
where K(St) =P
∫ 21hhρv . v dv is the kinetic energy for the material region occupied by P in the
present configuration κ .
We define an elastic body as one whose response depends on deformation gradient or on an
appropriate measure of strain. We assume the existence of a strainiiiii energyiiiiii or a stored energy per
unit mass ψ such that
tkivk,i = ρ ψ.
, (3.7)
and we define the (total) strain energy for each part P in the present configuration by
U(S t) =P
∫ ρ ψ dv . (3.8)
Then, by (3.6), (3.7), (3.8) the conservation of mass and the transport theorem we have
R(St) =dtdhhh[K(St) + U(St)] . (3.9)
We have thus proved the following theoremiiiiiii: The rate of work by contact and body forces is
equal to the sum of the rate of kinetic energy and the rate of the strain energy.
We now return to (3.7) and for an elastic body assume that the strain energy density
ψ = ψ′(F) [or ψ = ψ′(xi,A)].
Alternatively, we may assume that ψ depends on a measure of strain such as E and write
ψ = ψhh
(E) or ψ = ψhh
(EAB) . (3.10)
Now the material derivative of ψ is
ψ.
=∂EAB
∂ψhh
hhhhhE.
ABxi,B , (3.11)
where in obtaining (3.11) we have used (from Part I) the expressions
2EAB = CAB − δAB , CAB = xi,Axi,B ,
E.
AB = dikxk,Axi,B , C.
AB = 2E.
AB .
- 82 -
Introducing (3.11) into (3.7), we obtain
tkidki = ρ∂EAB
∂ψhh
hhhhh
or
(tki − ρ∂EAB
∂ψhh
hhhhhxi,Axk,B)dki = 0 , (3.12)
which must hold for all dki. Hence, we conclude that
tki = ρxi,Axk,B ∂EAB
∂ψhh
hhhhh or T = ρF∂E∂ψhh
hhhFT . (3.13)
If instead of the constitutive assumption (3.10), we begin with the alternative assumption
ψ = ψ(C) or ψ = ψ(CAB) , (3.14)
then in a manner analogous to the development that led to (3.14) we deduce that
tki = 2ρ xi,Axk,B ∂CAB
∂ψhhhhhh or T = 2ρ F∂C∂ψhhhhFT . (3.15)
- 83 -
4. Elastic solids: Linear constitutive equations.
We recall that in the context of infinitesimal kinematics discussed previously (Part I, sec-
tion 7), a function f is said to be of 0(εn) as ε → 0 if there exists a nonnegative real constant C
such that || f || < Cεn as ε → 0. Moreover, for infinitesimal deformation gradients, the distinction
between Lagrangian and Eulerian descriptions disappears; and that, to within 0(ε2), the relative
displacement gradient can be written as
ui,j = eij + ωij , (4.1)
where
eij =21hh (ui,j + uj,i) = 0(ε) , ωij =
21hh (ui,j − uj,i) = 0(ε) ,
as ε → 0 . (4.2)
We further assume that all derivatives (with respect to both coordinates and time) of kinematical
quantities, e.g., ui, eij, ωij, are also of 0(ε) as ε → 0.
We now introduce additional assumptions in order to obtain a completeiiiiiiii linear theory.
Thus, we assume that the fixed reference configuration (which was identified with the initial
configuration) is "stress-free", i.e., in this configuration the stress vector t and the body force b
are zero. We also assume that the body force b (or its components bi) and the stress vector t (or
its components ti) -- when expressed in suitable non-dimensional form -- as well as the com-
ponents of the stress tensor tki, all are of 0(ε) as ε → 0.
Recalling that for infinitesimal motion (see Eq. (7.32) of Part 1, section 7)
J−1 = 1 − eii = 1 − 0(ε) as ε → 0 , (4.3)
from the referential form of the conservation of mass, namely ρJ = ρo, we conclude that
ρ = ρo − 0(ε) as ε → 0 . (4.4)
Keeping in mind that t, b, u..
are all of 0(ε) as ε → 0, with the help of (4.4) and after the neglect
- 84 -
of terms of 0(ε2) as ε → 0, the equations of motion in the linear theory reduce to
tki,i + ρobk = ρov.k , (4.5)
tki = tik , (4.6)
where vi = u.i is the particle velocity, a comma denotes partial differentiation and the v
.i on the
right-hand side of (4.5) stands for∂t
∂vihhhh . It should be observed that each term in (4.5) is of 0(ε) as
ε → 0.
The equations of motion of the linear theory can also be obtained from the results of the
alternative derivation (see Part II, section 5). Thus, in terms of the non-symmetric Piola-
Kirchhoff stress tensor piA, the equations of motion are
piA,A + ρobi = ρov.i , (4.7)
piAxj,A = pjAxi,A , (4.8)
and we also recall the relations
piB = xi,AsAB , sAB = sBA , (4.9)
Jtki = xk,ApiA = xk,Axi,BsAB . (4.10)
Since J = 1 + 0(ε), xi,A = δiA + ui,A = δiA + 0(ε), XA,i = δAi + 0(ε) as ε → 0, we have
piA = JtkiXA,k
= tkiδAk + 0(ε2) as ε → 0 . (4.11)
Similarly, since now piA = 0(ε) as ε → 0, (4.9)1 after using (4.11) yields
sAB = piBXA,i
= piBδAi + 0(ε2) as ε → 0 . (4.12)
It follows from (4.11) and (4.12) that if terms of 0(ε2) as ε → 0 are neglected, then piA, sAB, tij
- 85 -
are all equal and the distinction between the Cauchy stress and the Piola-Kirchhoff stresses
disappears. Hence, the linearized version of the equations of motion (4.7) can also be written as
sAB,B + ρobA = ρou..
A . (4.13)
We now turn our attention to the linearized version of the constitutive equation discussed in
the previous section. We recall that in the context of nonlinear elasticity, the constitutive equa-
tions for the stress tensor may be expressed as
tki = ρxi,Axk,B ∂EAB
∂ψhh
hhhhh . (4.14)
In the linear theory discussed above the components of the stress tensor tki = 0(ε) as ε → 0.
Since the right-hand side of (4.14) must also be of 0(ε) as ε → 0 in the linear theory, for an elas-
tic body which is initially "stress-free", after linearization of (4.14), it will suffice to assume that
ψhh
is quadratic in the infinitesimal strain eij. To justify this stipulation, we proceed to obtain the
linearized version of (4.14). Recall that in the infinitesimal theory,
J = 1 + 0(ε)
xi,A = δiA + 0(ε)
eij = EABδAiδBj + 0(ε2) .
Using these expressions, we may linearize equation (4.14) to obtain
tki = ρxi,Axk,B dEAB
dψhh
hhhhh = ρoIL1 − 0(ε) MO
ILδiA + 0(ε) MO
ILδkB + 0(ε) MO dEAB
dψhh
hhhhh
= ρo dEAB
dψhh
hhhhh δiAδkB + 0(ε2) = ρo ∂emn
∂ψhh
hhhhh∂EAB
∂emnhhhhh δiAδkB + 0(ε2)
= ρo ∂emn
∂ψhh
hhhhh∂EAB
∂(EPQδPmδQn)hhhhhhhhhhhh δiAδkB
= ρo ∂emn
∂ψhh
hhhhh∂EAB
∂EPQhhhhh δPmδQnδiAδkB . (4.15)
- 86 -
Since EAB is a symmetric second order tensor, we have
∂EAB
∂EPQhhhhh =21hh
IJL ∂EAB
∂EPQhhhhh +∂EBA
∂EPQhhhhhMJO
=21hh I
LδPAδQB + δPBδQAMO
- 107 -
Supplements to the Main Text
ME 185 Class Notes
- 108 -
Supplements to the Main text in Class Notes
Rather than writing expressions in terms of components of vectors and tensors, it is some-
times convenient to use a coordinate free notation. In coordinate free notation, a vector is
referred to by the symbol v, whereas in indicial notation this vector is written in terms of its
components vi with respect to a basis ei. The values of the components vi are dependent on the
choice of basis. Similarly, a tensor may be referred to in coordinate free notation as T or in indi-
cial notation as Tij with respect to a given basis. Notice that tensor multiplication is not commu-
tative; however, the components of a tensor in indicial notation are commutative. The following
relationships between indicial and coordinate free notation are sometimes useful:
Coordinate Free Notation Indicial Notation
________________________ _________________
v (a vector) vi
A (a second order tensor) Aij
Av Aijvj
ATv Ajivj
AB AijBjk
ATB AjiBjk
ABT AijBkj
Examples from kinematics:
C = FTF CAB = FiAFiB
B = FFT Bij = FiAFjA
U2 = C UADUDB = CAB
Eigenvalues, eigenvectors, and characteristic equations.
In general, a second order tensor T operating on a vector v results in another vector (say) w
which is not necessarily in the same direction and does not necessarily have the same magnitude
- 109 -
as v. This operation is expressed symbolically as
Tv = w or Tijvj = wi . (1)
However, if a nonzero vector v is such that when T operates on it the resulting vector w is paral-
lel to v, then v is called an eigenvector of the tensor T.
In this case, we can write
Tv = βv or Tijvj = βvi , (2)
where the scalar β is called an eigenvalue corresponding to the eigenvector v. From (2), we can
write
(T − βI)v = 0 or (Tij − βδij)vj = 0 , (3)
where I is the identity tensor. Equation (3) represents three equations which must be satisfied by
all three components of v in order that v be an eigenvector of the tensor T. According to a
theorem of linear algebra, a homogeneous system of three equations for three unknowns (such as
equation (3)), has a non-trivial solution if and only if the determinant of the matrix of
coefficients vanishes, i.e. if
det(T − βI) = 0 or det(Tij − βδij) = 0 . (4)
Equation (4) represents a cubic equation in β, which may be written as
φ(β) = − β3 + I1β2 − I2β + I3 = 0 . (5)
Equation (5) is called the characteristic equation of T and I1 , I2 , and I3 are called the principal
invariants of T. Expressions for the principal invariants can be obtained by solving the system
(4), giving
- 110 -
I1 = trT I1 = Tii
I2 =21hh(tr2T − trT2) or I2 =
21hh(TiiTjj − TijTji) (6)
I3 = detT I3 = det(Tij) =61hhεijkεlmnTlTjmTkn .
In general equation (5) has three roots which are not necessarily real and are not necessarily dis-
tinct. However, in the special case where T is symmetric, we can prove some additional
theorems.
Theorem 1: If T is symmetric, then all the roots of the characteristic equation (5) are real.
Proof: We will prove this theorem by contradiction. Suppose first that the roots of (5) are not
real. According to a theorem of algebra, the complex roots of a polynomial with real
coefficients must occur in conjugate pairs. Therefore, if there exists a complex root β(1) of
(5), then there must also exist a root β(2) of (5) which is a conjugate of β(1). Thus β(1) and
β(2) must be of the form
β(1) = µ + iγβ(2) = µ − iγ , (7)
where i2 = −1. The eigenvectors which correspond to the eigenvalues β(1) and β(2) are of
the form
v(1) = αα + iδδ vj(1) = αj + iδj
or (8)
v(2) = αα − iδδ vj(2) = αj − iδj .
Since equation (2) must hold for all eigenvectors and their corresponding eigenvalues, we
can write
- 111 -
Tv(1) = β(1)v(1) Tijvj(1) = β(1)vi
(1)
or (9)
Tv(2) = β(2)v(2) Tijvj(2) = β(2)vi
(2) .
Multiply (9)1 by vi(2) and (9)2 by vi
(1) to obtain (in indicial notation)
β(1)vi(1)vi
(2) = Tijvj(1)vi
(2)
β(2)vi(2)vi
(1) = Tijvj(2)vi
(1) = Tjivj(2)vi
(1) = Tijvj(1)vi
(2) . (10)
In obtaining (10)2, we have made use of the facts that Tij is symmetric and that i and j
are dummy variables. Subtracting (10)2 from (10)1 gives
(β(1) − β(2))vi(1)vi
(2) = 0 . (11)
Substituting (7) and (8) into (11), we can write
0 = 2γi(αj + iδj) (αj − iδj) = 2γi(αjαj + δj δj) . (12)
The term in parenthesis in (12) is always greater than zero since it is the sum of two
squared real numbers. Therefore, we must have γ = 0. Equation (7) then implies that
β(1) and β(2) are both real, which is a contradiction of our initial assumption. We can
therefore conclude that if T is symmetric, all the roots of its characteristic equation are
real.
Theorem 2: The eigenvectors which correspond to distinct eigenvalues of a symmetric tensor
are orthogonal.
Proof: Suppose that v(1) and v(2) are eigenvectors of a symmetric tensor T corresponding to dis-
tinct eigenvalues β(1) and β(2), i.e. β(1) ≠ β(2). As in the previous proof, we can show that
- 112 -
equation (11) must be satisfied. Since β(1) ≠ β(2), we must have
v(1) . v(2) = 0 or vi(1)vi
(2) = 0 . (13)
Equation (13) is exactly the definition of orthogonal vectors, thus the eigenvectors v(1)
and v(2) are orthogonal.
Theorem 3: Every second order symmetric tensor has three linearly independent principal
directions.
Proof: For the proof of this theorem, see Advanced Engineering Mathematicsiiiiiiiiiiiiiiiiiiiiiiiiiiiiii, Wylie, p. 541.
Remark: In the preceding development, we can, without loss of generality, replace the eigen-
vector v with a unit vector m, where
m = v/ | v | . (14)
The vector m is called the normalized eigenvectoriiiiiiiiiiiiiiiiiii.
Using theorems 1, 2 and 3 and the previous remark, we can state that any symmetric tensor T
has three real eigenvalues β(1) , β(2) , and β(3) and three orthonormal eigenvectors
m(1) , m(2) , and m(3). It can be shown, using a theorem of linear algebra, that since T has three
linearly independent eigenvectors, it can be transformed into a diagonalized form, denoted by
Λ, using the transformation
T′ij = Λij = amianjTmn ,
where
aij = ei. m(j) and xi = aijx′j . (15)
Here the primed quantities correspond to the transformed system. It can also be shown that the
- 113 -
diagonal members of Λ are simply the eigenvalues, so that
(Λij) =
RJJJQ0
0
β(1)
0
β(2)
0
β(3)
0
0HJJJP
. (16)
Quadratic forms and positive definiteness.
Associated with any symmetric second order tensor T is a scalar valued function of a vector
Q(v), where v is an arbitrary vector, defined by
Q(v) = (Tv) . v = Tij vjvi = v . (Tv) . (17)
We call Q(v) a quadratic formiiiiiiiiiiii. The tensor T is said to be positive definiteiiiiiiiiiiiii if for all
v ≠ 0 , Q(v) > 0 .
Theorem 4: A symmetric second order tensor T is positive definite if and only if all the eigen-
values of T are positive.
Proof: Again using a theorem from linear algebra, one can conclude that since the eigenvectors
m(1) , m(2) , and m(3) are orthogonal in a 3-dimensional space, then any arbitrary vector w
in that space may be represented as a linear combination of m(1) , m(2) , and m(3). Thus, we
can write
w = a1m(1) + a2m(2) + a3m(3) . (18)
The eigenvalues of T corresponding to m(1) , m(2) , and m(3) are β(1) , β(2) , and β(3), as
before, so equation (2) gives
Tm(i) = β(i)m(i) . (19)
- 114 -
Using (18) and (19), we can write
Tw = a1β(1)m(1) + a2β(2)m(2) + a3β(3)m(3) . (20)
Substituting (20) into (17) gives
Q(w) = Tijwjwi = a12β(1) + a
22β(2) + a
32β(3) . (21)
Necessity: If T is positive definite, then Q(w) > 0 for all w. Since a1 , a2 , and a3 are all real
numbers, the square of each of these terms must be positive (or zero). If w = a1m(1), then
Q(w) > 0 implies from (21) that β(1)al2 > 0, or that β(1) > 0. Since w is arbitrary, a similar
argument shows that β(2) and β(3) are both greater than zero. Thus, if T is positive definite,
then all the eigenvalues of T are positive.
Sufficiency: If all the eigenvalues of T are greater than zero, then by (21) it follows that
Q(w) > 0 for all w ≠ 0. Therefore, T is positive definite if all the eigenvalues of T are posi-
tive.
Theorem 5: If B is a positive definite symmetric tensor, then there exists a unique symmetric
positive definite square root T of B such that T2 = B.
Remarks on the calculation of the square root of B:
Let Bik = aipakqΛpq (see (15)), where Λ is a diagonal matrix as in equation (16). Define a
tensor Λ 21hh
pq as
(Λ 21hh
)pq =
RJJJQ0
0
√dddβ(1)
0
√dddβ(2)
0
√dddβ(3)
0
0HJJJP
,
so that
- 115 -
Λ 21hh
pqΛ 21hh
qs = Λps . (22)
Defining Tij = aipajqΛ 21hh
pq, we can show that
TijTjk = aipajq Λ 21hh
pqajmaknΛ 21hh
mn = aipajqajmaknΛ 21hh
pqΛ 21hh
mn
= aipaknδqm Λ 21hh
pqΛ 21hh
mn = aipaknΛpn = Bik
or
T2 = B . (23)
Theorem 6: Polar decomposition theorem (see class notes section I/3)
Any non-singular (invertible) tensor F may be uniquely expressed as
F = RU = VR , (24)
where R is orthogonal and U and V are symmetric positive definite tensors.
Proof: Existence
Define U = (FTF) 21hh
and R = FU−1. We first show that U is symmetric postive definite.
To show that U2 is positive definite, consider the quadratic form
Q(v) = v . U2v = v . FTFv = Fv . Fv .
Since F is invertible, then Fv ≠ 0 if v ≠ 0. Assuming that F and v are both real, it fol-
lows that Q(v) > 0 if v = 0. Thus, U2 is positive definite. Also, since FTF is sym-
metric, then U2 must be symmetric. From theorem 5, we know that U2 must have a
unique positive definite square root U, where U is also symmetric. To show that R is
orthogonal, note that
RTR = (U−1)TFTFU−1 = U−1FTFU−1 = U−1UUU−1 = I .
- 116 -
Using the definitions of U and R, we can write
RU = FU−1U = F .
- 117 -
Thus, we have shown the existence of the decomposition (24).
Uniqueness
To show uniqueness of the decomposition (24), assume first that two such decomposi-
tions exist, so that F = RU = R*U*. The star is used here to denote an alternative
decomposition. Notice that
FTF = UTRTRU = U2
and
FTF = U*TR*TR*U* = U*2 ,
thus U2 = U*2. By theorem 5, the square roots of U2 and U*2 are unique, so we must
have U = U*. Notice that
0 = F − F = RU − R*U* = (R − R*)U .
Since U is non-singular, U−1 exists. Thus,
0 = (R − R*)UU−1 = R − R* ,
and so R = R*. Thus, the decomposition F = RU is unique. Similarly, one can prove
the existence and uniqueness of the decomposition F = VRhh
, where we temporarily
allow Rhh
to be different from R. The relationship between V, U, R, and Rhh
may be
determined by observing that
F = RU = (RURT)R = VRhh
,
where RURT is symmetric and positive definite. Since the decomposition of F is
unique, we conclude that
V = RURT , Rhh
= R , and U = RTVR . (25)
- 118 -
Extremal properties of quadratic forms.
Consider the symmetric tensor T and the quadratic form
Q(v) = v . Tv = Tijvjvi . (26)
We wish to know the extreme values of Q(v) subject to the constraint that v be a unit vector
(i.e. v . v = vi vi = 1). Using the method of Lagrange multipliers, the conditions for the
extremum of (26) are
∂vk
∂hhhh Tijvivj − β(vivi − 1) = 0 , (27)
where β is the undetermined multiplier and v . v − 1 = 0 is the constraint equation. Performing
the differentiation, (27) becomes
Tijδikvj + Tijviδjk −2βviδik = 0
or
Tkjvj + Tikvi − 2βvk = 0 . (28)
Since T is symmetric, (28) becomes
Tkjvj − βvk = 0 . (29)
Therefore, Q(v) attains extreme values when v is a eigenvector of T. Using (2), it is clear that if
m is a unit eigenvector of T with corresponding eigenvalue β, then
Q(m) = m . (Tm) = β . (30)
In summary, the quadratic form obtains its extremum values when it is operating on an eigenvec-
tor. If the eigenvectors are normal, these extremum values are simply the corresponding eigen-
values.
- 119 -
Relationship between eigenvalues of U, C, V, B, E, and D and the principal stretch.
(see class notes sections I/4 and I/5)
The principal directions are the directions in which the stretch attains an extreme value.
The principal stretch λ is the extreme value of stretch in a given principal direction. As derived
in section I/4, the eigenvalues of U and V are equal to the principal values of stretch and the
eigenvalues of C and B are equal to λ2. It was also shown in this section that the eigenvectors of
U and C are coincident and that the eigenvectors of V and B coincide.
The relative strain tensor E is given by
E =21hh(C − I) . (31)
The principal values of strain, denoted by β, are equal to the eigenvalues of E. The principal
directions of strain, denoted by m, are equal to the eigenvectors of E. From (2), we have
Em = βm . (32)
Substituting (31) into (32) gives
21hh(C − I)m = βm ,
or
Cm = (2β + 1)m . (33)
Equation (33) implies that m is an eigenvector of C as well as of E and that 2β + 1 is an eigen-
value of C. As previously noted, the eigenvalue of C is λ2, so
2β + 1 = λ2 or β =21hh(λ2 − 1) . (34)
- 120 -
Thus, the eigenvectors of U , C , and E coincide and the principal value of strain is related to the
principal value of stretch by equation (34).
As was derived in section I/5, the eigenvalues and eigenvectors of the rate of deformation
tensor D are given by
Dm =λλ.hhm =
dtdhhh(ln λ) m , (35)
where m is an eigenvector of D, (m is also called the "principal direction of stretch"). It is clear
from the form of (35) that λ.
/ λ (called the logarithmic rate of stretching) is an eigenvalue of D.
Example calculation
Suppose that at some instant at a particular point in the body, the deformation gradient ten-
sor F is given by
[FiA] =RJJQ01
2
0
2
1
1
0
0 HJJP
. (36)
Notice that (36) yields a symmetric positive definite tensor. Since the polar decomposition of F
is unique, it follows that V = U = F and R = I . The principal invariants of F = U are
I1 = tr U = UAA = 5
I2 =21hh(tr2U − trU2) =
21hh(UAAUBB − UABUBA) = 7 (37)
I3 = det U = 3 .
The characteristic equation of F is then
−β3 + 5β2 − 7β + 3 = 0 . (38)
The roots of (38), i.e. the eigenvalues of F, are
- 121 -
β(1) = 3
β(2) = 1 (39)
β(3) = 1 .
To solve for the eigenvectors of F, substitute each of the eigenvalues into equation (2), or
Fv = βv . (40)
For β(1) = 3, equation (40) gives
(F − 3I)M(1) = 0 ,
or
− M1(1) + M
2(1) = 0 and −2M
3(1) = 0 .
Thus, the eigenvector M(1) is given by
[Mi(1)] = c1
RJJQ01
1 HJJP
,
where c1 is an arbitrary constant. Similarly, for β(2) = β(3) = 1, equation (40) gives
(F − I)M(α) = 0
where α = 2,3, or
M1(α) = −M
2(α) .
Solving for M(α) gives
[Mi(2)] = c2
RJJQ 0
1
−1 HJJP
and [Mi(3)] = c3
RJJQ10
0 HJJP
.
We may take unit eigenvectors
- 122 -
[Mi(1)]=
√dd2
1hhhRJJQ01
1 HJJP
, [Mi(2)] =
√dd2
1hhhRJJQ 0
1
−1 HJJP
, [Mi(3)] =
RJJQ10
0 HJJP
,
which form a right-handed triad. Note that every vector in the space spanned by M(2) and M(3)
is also an eigenvector with eigenvalue 1.
One can better understand the physical significance of these results by looking at the effect
of the deformation (36) upon a unit cube. The graph below gives the cross section of the cube
before and after the deformation. As expected from the preceding calculations, the principal
directions (i.e. the directions in which the stretch attains extremum values) are in the directions
of the eigenvectors M(1) , M(2) , and M(3). Line elements along M(1) are stretched to three
times the initial length. Thus, the stretch λ = 3 is equal to β(1), as expected. Line elements in the
directions M(2) and M(3) retain their initial length. Therefore, the stretch λ in these directions
equals unity, corresponding to β(2) = β(3) = 1. In this example, line elements along principal
directions undergo pure stretch as a result of the deformation; however, line elements in other
directions may undergo both stretch and rotation. This characteristic is true in general whenever
the deformation F is of the pure stretch type (i.e. whenever R = I).
- 123 -
Supplement to Part I, Section 5
Material line elements, area elements, volume elements and their material time derivatives.
Preliminary results
Claim:
det A =61hh εijkεlmnAliAmjAnk (1)
Proof: Consider an arbitrary tensor A with components Aij. Let am = Am1, bm = Am2, and
cm = Am3. Recall that
a . (b × c) = εlmnalbmcn = εlmnAl1Am2An3 = det
RJJQA13
A12
A11
A23
A22
A21
A33
A32
A31HJJP
= det AT ≡ A . (2)
Now define Tijk by
Tijk = εlmnAliAmjAnk (3)
where by (2), T123 = A. Since l, m, and n are dummy indices, we may switch them and use
the properties of εlmn to show that
Tijk = Tjki = Tkij = − Tikj = − Tkji = − Tjik . (4)
Recalling the properties of εijk, we can conclude from (4) that Tijk must be a multiple of εijk.
>From (2) and (4), we see that
T231 = T312 = T123 = A
T321 = T213 = T132 = − A
Tijk = 0 if i = j , j = k , or i = k .
Hence, it is clear that
Tijk = A εijk . (5)
- 124 -
Equate (3) and (5) and multiply the resulting equation by61hh εijk, noting that εijkεijk = 6, to
get
A = det AT =61hh εijkεlmnAliAmjAnk . (6)
Since det AT = det A for any tensor A, we have proved equation (1).
Claim:
∂A∂(det A)hhhhhhhh = (det A) A−T (7)
Proof: Assume that the components of A are independent, so that we can write
∂Ars
∂Aijhhhhh = δirδjs . (8)
Now differentiate (6) with respect to Ars to obtain
∂Ars
∂(det A)hhhhhhhh =∂Ars
∂Ahhhhh =61hh εijkεlmn δrlδsiAmjAnk + AliδrmδsjAnk + AliAmjδrnδsk
=61hhεsjkεrmnAmjAnk + εiskεlrnAliAnk + εijsεlmrAliAmj
=61hh3εsjkεrmnAmjAnk =
21hhεsjkεrmnAmjAnk . (9)
Suppose now that A is non-singular (det A ≠ 0), so that there exists a tensor B with com-
ponents Bij such that AliBir = δlr. If this is the case, then (9) may be rewritten using (3) and
(5) as
∂Ars
∂Ahhhhh =21hh εsjk(εlmnδrl)AmjAnk
=21hh εsjkεlmn(AliBir)AmjAnk
=21hh εsjk(εlmnAliAmjAnk)Bir
=21hh εsjkεijkABir =
21hh(2ABsr) = ABsr . (10)
- 125 -
Denoting B as A−T, equation (10) becomes
∂A∂(det A)hhhhhhhh = A A−T = (det A) A−T ,
which is identical to (7).
Material line elements
Consider a motion of a body B given by x = χ (X , t), where X is the position vector of a
certain particle in the reference configuration and x is the position vector of the same particle in
the current configuration at time t. Two neighboring material particles in the reference
configuration are located at X and X + dX. Consider a material line elementiiiiiiiiiiiiiiiii dX in the reference
configuration at the point X and having length dS and direction M, so that dX = dS M. Under
the motion χ(X , t), the material particles X and X + dX are taken to the positions x and x + dx.
The corresponding material line element dx in the current configuration having length ds and
direction m is related to dX by
dx = F dX , or dxi = FiA dXA = xi,AdXA , (11)
where FiA = xi,A are the components of the deformation gradient F. A diagram of this motion of
the material line segment is given below. Recall from Section I/4 of the class notes that the
square of the length of a material line element in the current configuration is given by
ds2 = xi,Axi,BdXAdXB . (12)
Recalling that xi,A
hhh.= vi,jxj,A, we can take the material derivative of (12) to get
ds2hhh.
= xi,A
hhh.xi,BdXAdXB + xi,Axi,B
hhh.dXAdXB
= vi,jxj,Axi,BdXAdXB + xi,Avi,jxj,BdXAdXB
= (vi,j + vj,i)xi,Axj,BdXAdXB
= 2dijxi,Axj,BdXAdXB . (13)
- 126 -
This last step uses the definition of the rate of deformation tensor D =21hh(L + LT). Equation (13)
may be rewritten using (11) as
ds2hhh.
= 2dijdxidxj . (14)
Material area elements
Let 1dX and 2dX represent two material line elements in the reference configuration located
at X. Suppose that the motion χ (X ,t) takes 1dX, 2dX into the elements 1dx, 2dx, respectively, at
x in the current configuration. Define the material area elementiiiiiiiiiiiiiiiiii dA with orientation N in the
reference configuration by
N dA = 1dX × 2dX (15)
so that
dAL = NLdA = N dA . eL = (1dX × 2dX) . eL = εLMN1dXM
2dXN . (16)
The corresponding material area element in the current configuration is da with orientation n
(see Figure 2) defined by
n da = 1dx × 2dx , (17)
so that
ni da = n da . ei = (1dx × 2dx) . ei (18)
>From (18) and (11) and noting that xm,DXD,i = δmi, we can write
ni da = εijkxj,Axk,B1dXA
2dXB = εmjkδmixj,Axk,B1dXA
2dXB
= εmjk(xm,DXD,i)xj,Axk,B1dXA
2dXB
= εmjkxm,Dxj,Axk,BXD,i1dXA
2dXB .
Using (3), (5), and (16), this expression becomes
ni da = εDAB(det F)XD,i1dXA
2dXB
- 127 -
or
ni da = J XD,i ND dA . (19)
Assuming that F is non-singular and using (7) and (19), we can write
ni dahhhh.
= J.
XD,i ND dA + J XD,i
hhhh.ND dA . (20)
Notice that
J.
=dtdJhhh =
∂FjA
∂Jhhhhh∂t
∂FjAhhhhh = J (F−1)Aj vj,kFkA = J vj,kδkj = J vj,j . (21)
Using (21), equation (20) becomes
ni dahhhh.
= (J XD,ivj,j − J XD,jvj,i)ND dA
or
ni dahhhh.
= J(vj,jXD,i − vj,iXD,j)ND dA . (22)
Material volume elements
Let 1dX, 2dX, and 3dX be three material line elements forming a right handed system
located at X in the reference configuration. Suppose that the motion χ (X , t) takes 1dX, 2dX,
and 3dX into three line elements 1dx, 2dx, and 3dx, respectively, at x in the current configuration
(see Figure 3). Define the material volume elementiiiiiiiiiiiiiiiiiiii dV in the reference configuration by
dV = 1dX . (2dX × 3dX) = εABC1dXA
2dXB3dXC . (23)
The corresponding material volume element in the current configuration is then
dv = 1dx . (2dx × 3dx) = εijk1dxi
2dxj3dxk . (24)
Using equations (3), (5), (11), and (23), we can write (24) as
dv = εijkxi,Axj,Bxk,C1dXA
2dXB3dXC = εABC(det F) 1dXA
2dXB3dXC
- 128 -
= J(εABC1dXA
2dXB3dXC)
- 129 -
or
dv = J dV . (25)
Taking the material derivative of (25) gives
dvhh.
= J.
dV = J vj,jdV = J(div v)dV , (26)
where equation (21) is used for the calculation of J..
A special case of the above analysis occurs for isochoric motionsiiiiiiiiiiiiiii (i.e. motions for which
dv = dV for all volumes occupied by any material region of the body). It is clear from this
definition that dvhh.
= 0 for isochoric motions. We see from equations (25) and (26) that the neces-
sary and sufficient condition for isochoric motion may be expressed either as
J = 1 or vi,i = div v = 0 . (27)
Summary
A summary of the basic equations derived in this appendix are given in the following:
Material line, area, and volume elements:
ds2 = xi,Axi,BdXAdXB
ni da = J XD,i ND dA
dv = J dV
Material derivatives of line, area, and volume elements:
ds2hhh.
= 2 dijxi,Axj,BdXAdXB
dai
hhh.= J(vj,jXD,i − vj,iXD,j)dAD
dvhh.
= J vj,jdV .
- 130 -
Supplement to Part I, Section 7
Interpretation of infinitesimal strain measures.
Recall the following expressions:
ds2 = dxidxi = xi,Axi,BdXAdXB
dS2 = dXAdXA = XA,iXA,jdxidxj
EAB =21hh(CAB − δAB) =
21hh(UA,B + UB,A + UM,AUM,B)
eij =21hh(δij − cij) =
21hh(ui,j + uj,i − um,ium,j) . (1)
Also recall that the stretch λ and the extension E are defined by
λ =dSdshhh = √ddddddddCABMAMB = √dddddddddddd1 + 2 EABMAMB (2a)
= 1/√dddddcijmimj = 1/√ddddddddd1 − 2eijmimj
and
E =dS
ds − dShhhhhhh = λ − 1 , (2b)
where
MAMA = mimi = 1 and dXA = MAdS , dxi = mids . (3)
For an infinitesimal deformation (see class notes Part I/7), we may approximate EAB by neglect-
ing term of O(ε2) as ε → 0 as
EAB =21hh(UA,B + UB,A) = O(ε) as ε → 0 . (4)
The distinction between Lagrangian and Eulerian strain disappears for an infinitesimal deforma-
tion. Thus,
- 131 -
eij = EABδiAδjB = O(ε) as ε → 0 . (5)
Hence, for an infinitesimal deformation the extension and the stretch given in (2) may be written
as
E = λ − 1 = EABMAMB + O(ε2) = eijmimj + O(ε2) (6)
and
λ = √dddddddddddd1 + 2EABMAMB = 1 + EABMAMB + O(ε2) (7a)
or, alternatively,
λ = 1/√ddddddddd1 − 2eijmimj = 1 + eijmimj + O(ε2) . (7b)
In the following discussion, we will use e11, e22, ..., to denote the components of the infinitesimal
strain tensor, since from (5) no distinction needs to be made between Lagrangian and Eulerian
strain as ε → 0. Consider a line element which lies along the X1 axis, i.e.
dX1 = dS , dX2 = dX3 = 0 , and MA = (1, 0, 0) ,
so that from (6)
E =dS
ds − dShhhhhhh = e11 . (8)
That is, e11 represents the extension (the change in length divided by the original length) of a
line element lying along the X1 axis. We note that to within O(ε2), the direction mi in the current
configuration coincides with the direction MA in the reference configuration. Similar results
hold for e22 and e33. In summary: the diagonal elements of the infinitesimal strain tensor
represent the extension of line elements in the corresponding coordinate directions.
Consider two orthogonal elements which initially lie along the X1 and X2 axes, so that
dX1 = dS , dX2 = dX3 = 0 , MA = (1, 0, 0) (9)
dXhh
2 = dSh
, dXhh
1 = dXhh
3 = 0 , Mhh
A = (0, 1, 0) .
- 132 -
Let θ denote the angle between the corresponding deformed line elements dx and dxh, whose
lengths are ds and dsh. >From (8), neglecting terms of O(ε2) as ε → 0, we have
ds = (1 + e11)dS and dsh
= (1 + e22)dSh
. (10)
Now observe that
cos θ =| dx | | dx
h|
dx . dxh
hhhhhhhhhh =ds ds
hdxi dx
hihhhhhh = xi,Axi,B
ds dsh
dXAdXhh
Bhhhhhhhh
= CABds ds
hdXAdX
hhBhhhhhhhh = CAB
(1 + e11)(1 + e22)
MAMhh
Bhhhhhhhhhhhhhhh
= C12/(1 + e11)(1 + e22) = 2e12/(1 + e11 + e22) + O(ε2)
= 2e12 + O(ε2) , (11)
where we have used (1)3, (5), (9), and (10) in deriving (11).
Let φ be the amount θ differs from a right angle (i.e. φ = −θ + π/2), so that
cos θ = sin φ = φ + O(φ3) . (12)
>From (11) and (12), we see, after neglecting terms of O(ε2), that
φ = 2e12 or e12 =2φhh . (13)
The angle φ is known as the sheariiiii angleiiiii. Similar results hold for e13 and e23. Hence: the off-
diagonal components of the infinitesimal strain tensor are seen to represent half of the change in
angle between two line elements initially along the corresponding coordinate axes.
Recall that dv=JdV, where J = det(xi,A). In the class notes Section I/7, it was shown that
neglecting terms of O(ε2) as ε → 0 gives
J = 1 + ekk . (14)
- 133 -
Thus, dv = JdV = dV + ekkdV, or
dVdv − dVhhhhhhhh = ekk . (15)
The invariant ekk is known as the dilationiiiiiii. Equation (15) implies that the trace of the
infinitesimal strain tensor measures the local change in volume per unit volume.
- 134 -
Supplement to Part III, Section 2
Isotropic tensors.
Under a rotation of coordinate system, recall that the components of a vector change
according to the rule
xi = aimx′m or x′m = ajmxj , (1)
where xi are the components of a vector x with respect to one coordinate system and x′m are the
components of the same vector with respect to a different coordinate system. The coefficients
aim are components of an orthogonal tensor such that
aijaik = ajiaki = δjk and det(aij) = ± 1 . (2)
Also recall that the components Tij...k of a tensor T transform according to the rule
Tij...k = aipajq. . . akrT′pq...r , (3)
where aij’s obey equation (2).
Definition: A tensor T is called isotropiciiiiiiii if its components are invariant to rotation of the coor-
dinate system. Thus, if T is isotropic, then
Tij...k = T′ij...k ,
or from (3),
Tij...k = aipajq. . . akrTpq...r (4)
for all orthogonal aij.
Theorem 1: A scalar invariant is an isotropic tensor of order zero.
Proof: Let φ = φ(xi) be a scalar invariant, so that from (3) φ(xi) = φ(x′i) = φ′. Since φ = φ′, the
scalar invariant φ satisfies the definition of an isotropic tensor given by (4).
Theorem 2: The only isotropic tensor of order one is the zero vector.
- 135 -
Proof: If ti are the components of an isotropic vector t, then from (4)
ti = ajitj (5)
for all components aji of orthogonal tensors. Consider, for instance, the choice
RQaji
HP =
RJJQ 0
0
−1
0
−1
0
1
0
0 HJJP
.
With this choice for aji, (5) becomes (t1 , t2 , t3) = (−t1 , −t2 , t3), which implies that t1 = 0
and t2 = 0. Now consider the choice
RQaji
HP =
RJJQ00
1
0
−1
0
−1
0
0 HJJP
.
for which equation (5) becomes (t1 , t2 , t3) = (t1 , −t2 , −t3), and so we must have t3 = 0
also. Thus, if t is an isotropic tensor of order one, then t = 0.
Theorem 3: Every isotropic tensor of order two is a scalar multiple of δij.
Proof: Let Tij be the components of an isotropic tensor T, so that from (4)
Tij = aipajqTpq (6)
for all orthogonal tensor components aij. Consider now the following choices for aij and the
resulting forms of equation (6):
RQaij
HP =
RJJQ 0
−1
0
1
0
0
0
0
−1 HJJP
so that
RJJJQT31
T21
T11
T32
T22
T12
T33
T23
T13HJJJP
=
RJJJQ−T23
T13
T33
−T21
T11
T31
T22
−T12
−T32HJJJP
,
and
RQaij
HP =
RJJQ01
0
−1
0
0
0
0
−1 HJJP
so that
RJJJQT31
T21
T11
T32
T22
T12
T33
T23
T13HJJJP
=
RJJJQ T23
−T13
T33
−T21
T11
−T31
T22
−T12
T32HJJJP
. (7)
- 136 -
In order for (7)1 and (7)2 to both be satisfied, we must have
T11 = T22 = T33 and T12 = T21 = T13 = T31 = T23 = T32 = 0 ,
or
Tij = λδij , (8)
where
λ = T11 = T22 = T33 .
Theorem 4: Every isotropic tensor of order three is a scalar multiple of the permutation tensor
εijk.
Proof: Let Tijk be an isotropic third order tensor, so that from (4) we can write
Tijk = aipajqakrTpqr (9)
for all orthogonal aij. We can represent Tijk as
RQTijk
HP =
RJJJQT311
T211
T111
T312
T212
T112
T313
T213|
T113|
T321
T221
T121
T322
T222
T122
T323
T223|
T123|
T331
T231
T131
T332
T232
T132
T333
T233
T133HJJJP
.
Consider the following choices for aij and the resulting forms of (9):
RQaij
HP=
RJJQ 0
−1
0
1
0
0
0
0
−1 HJJP
then RQTijk
HP=
RJJJQ T233
−T133
−T333
T231
−T131
−T331
−T232
T132|
T332|
T213
−T113
−T313
T211
−T111
−T311
−T212
T112|
T312|
−T223
T123
T323
−T221
T121
T321
T222
−T122
−T322HJJJP
RQaij
HP=
RJJQ01
0
1
0
0
0
0
1 HJJP
then RQTijk
HP=
RJJJQ T233
T133
T333
T231
T131
T331
T232
T132|
T332|
T213
T113
T313
T211
T111
T311
T212
T112|
T312|
T223
T123
T323
T221
T121
T321
T222
T122
T322HJJJP
RQaij
HP=
RJJQ−1
0
0
0
1
0
0
0
1 HJJP
then RQTijk
HP=
RJJJQ−T133
T233
T333
−T132
T232
T332
T131
−T231|
−T331|
−T123
T223
T323
−T122
T222
T322
T121
−T221|
−T321|
T113
−T213
−T313
T112
−T212
−T312
−T111
T211
T311HJJJP
.
(10)
- 137 -
>From equation (10), we must have
T333 = T331 = T332 = T111 = T112
= T122 = T133 = T211 = T233 = T311 = T322 = T121 = T131 = T212 == T232 = T313 = T323 = 0
and
T123 = T231 = T312 = −T321 = −T132 = −T213 ≡ λ ,
thus we can write
Tijk = λεijk . (11)
Sufficiency may also be shown by substituting equation (11) into equation (3), giving
aipajqakrTpqr = aipajqakrλεpqr = λ(det apq)εijk
which is equivalent to equation (9).
Theorem 5: The most general fourth order isotropic tensor has the form
Tijkl = λδijδkl + µδikδjl + γδilδjk .
Proof: Let Tijkl be the components of an isotropic tensor, so that from (4)
Tijkl = aipajqakralsTpqrs (12)
for all orthogonal aij. We proceed as in the previous proofs, choosing aij to be
RQaij
HP =
RJJQ 0
0
−1
0
−1
0
1
0
0 HJJP
,
RJJQ00
1
0
−1
0
−1
0
0 HJJP
,
RJJQ01
0
1
0
0
0
0
1 HJJP
,
RJJQ10
0
0
1
0
0
0
−1 HJJP
.
Using the above choices for aij in equation (12), we can show that the non-zero values of
Tijkl are related as follows:
T1111 = T2222 = T3333
T1122 = T2211 = T1133 = T3311 = T3322 = T2233 = λT1212 = T2121 = T1313 = T3131 = T2323 = T3232 = µT1221 = T2112 = T1331 = T3113 = T2332 = T3223 = γ . (13)
- 138 -
Now consider the choice
RQaij
HP =
RJJQ 0
−1/√dd21/√dd2
0
1/√dd21/√dd2
1
0
0 HJJP
.
If we substitute this value for aij into (12) and set i = j = k = l = 1, we find that
T1111 =41hh(T1111 + T2211 + T2121 + T1221 + T2112 + T1212 + T1122 + T2222) ,
or using (13),
T1111 = λ + µ + γ . (14)
>From the restriction (13) and (14), we see that Tijkl can be represented as
Tijkl = λδijδkl + µδikδjl + γδilδjk . (15)
Sufficiency may be shown by substituting the expression for Tijkl given by equation (15)
into equation (4), yielding
aipajqakralsTpqrs = aipajqakrals(λδpqδrs + µδprδqs + γδpsδqr)
= λaipajpakralr + µaipajqakpalq + γaipajqakqalp
= λδijδkl + µδikδjl + γδilδjk = Tijkl
which is identical to equation (12). If an isotropic fourth order tensor is restricted to have
the symmetry
Tijkl = Tjikl (16)
(i.e. symmetry in the first two indices), then from equation (13) we see that γ = µ. Equation
(15) may thus be written for this special case as
Tijkl = λδijδkl + µ(δikδjl + δilδjk) . (17)
Equation (17) is the most general form for an isotropic fourth order tensor symmetric in i
and j. Notice that the representation given in equation (17) also possesses the additional
symmetries given by
- 139 -
Tijkl = Tijlk = Tklij . (18)
- 140 -
Supplement to Part III, Section 4
Material symmetry of elastic solids (general considerations).
For a homogeneous elastic solid, the constitutive equation of interest expresses the strain
energy per unit mass as a function of some measure of strain. Thus, let the strain energy per unit
mass ψ be given by
ψ = ψ(EAB) . (1)
We examine the behavior of (1) under a change of coordinates in the reference configuration
given by
XP = APQX′Q ,
where
APQ = EP. E′Q and APM AQM = AMP AMQ = δPQ . (2)
Under the transformation (2), EAB becomes
EPQ = APM AQN E′MN , (3)
where E′MN are the components of the strain tensor referred to the base vectors e′A. Since ψ is a
scalar invariant, we know that
ψ′ = ψ . (4)
With the use of (1) and (3), the strain energy function after the transformation (2) can be written
as
ψ′ = ψ(APM AQN E′MN) , (5a)
where as before
ψ = ψ(EPQ) . (5b)
Substituting (5) into (4), we conclude that
ψ(EPQ) = ψ(APM AQN E′MN) (6)
- 141 -
where ψhh
is in general a different function than ψ. It is clear from equation (6) that the response
of an elastic solid depends, in general, on the coordinates used to describe the reference
configuration. The symmetry of a homogeneous elastic solid is characterized by the set of
orthogonal coordinate transformations which leave the strain energy function form-invariant;
that is, the set of APQ such that ψhh
is the same as ψ, or such that
ψ(EPQ) = ψ(E′PQ) . (7)
If equation (7) holds for all possible orthogonal coordinate transformations (i.e. for all orthogo-
nal tensors APQ), then the material is said to be isotropiciiiiiiii.
Linear elastic solids
It was shown in the class notes that, in the linear theory, the strain energy function for a
homogeneous elastic solid may be expressed in the form
ρoψ =21hh CABMNEABEMN , (8)
where CABMN is a fourth order tensor of material constants which has, in general, 81 indepen-
dent components. Since EAB is a symmetric tensor, the coefficients CABMN in (8) will have the
obvious symmetries
CABMN = CBAMN = CABNM . (9)
In addition, recalling that the strain energy is a scalar invariant we may write (after changing the
dummies AB into MN and MN into AB)
ρoψ =21hh CABMNEABEMN =
21hh CMNABEMNEAB . (10)
It then follows that CABMN must by symmetric in the pair of indices (AB) and (MN), i.e.,
CABMN = CMNAB . (11)
In summary, the fourth order tensor in (8) must satisfy the symmetries
CABMN = CBAMN = CABNM = CMNAB . (12)
- 142 -
The symmetry of CABMN shown in (12) allows us to reduce the number of independent com-
ponents of CABMN from 81 to 21. The remaining independent components are
C1111
C2222
C1122
C3333
C2233
C1133
C2323
C3323
C2223
C1123
C1313
C2313
C3313
C2213
C1113
C1212 .
C1312
C2312
C3312
C2212
C1112
(13)
Linear elastic solids with symmetry
As mentioned previously, the symmetry of an elastic solid is characterized by the set of
orthogonal tensors APQ such that (7) holds. Substituting (8) into (7), we find
ρoψ =21hh CABMNEABEMN =
21hh CABMNE′ABE′MN . (14)
Using (3), (14) can be written as
CABMNAAPABQAMRANSE′PQE′RS = CABMNE′ABE′MN . (15)
Switching dummy indices, (15) becomes
(CABMN − CPQRSAPAAQBARMASN)E′ABE′MN = 0 . (16)
Since (16) must hold for all E′AB and since the quantity in parenthesis in (16) is independent of
E′AB, we may conclude that
CABMN = CPQRSAPAAQBARMASN (17)
for all orthogonal APQ which lie in the set which characterizes the symmetry of the material.
Isotropic linear elastic solids
If the material is isotropic, then equation (17) must hold for all orthogonal APQ. Hence,
CABMN in an isotropic tensor which from (12) is symmetric in the first two indices. >From the
analysis given in the Appendix to Section 2 of Part III, we can represent CABMN as
- 143 -
CABMN = λδABδMN + µ(δAMδBN + δANδBM) . (18)
Eulerian strain
In the linear theory, there is no distinction between Lagrangian and Eulerian strain as is
clear from the expression
EAB = δAiδBjeij + O(ε2) .
Substituting this into (8), we may write
ρoψ =21hh (CABMNδAiδBjδMkδNl)eijekl + O(ε2) . (19)
Now define
Cijkl = CABMNδAiδBjδMkδNl , (20)
so that (19) becomes
ρoψ =21hh Cijkl eij ekl (21)
If the material is isotropic, then from (18)
Cijkl = λδijδkl + µ(δikδjl + δil δjk) . (22)
We will now consider materials with various symmetries and analyze the effect of the different
symmetries on CABMN.
Case I(a) - Symmetry with respect to the X1−X2 plane
Consider a homogeneous elastic solid which in the reference configuration has material
symmetry only with respect to the X1−X2 plane. For a material with such a symmetry, the
mechanical behavior in the X3 direction is the same as that in the −X3 direction (dentoed by X′3
in Figure 1). The coordinate transformation characterizing this particular type of symmetry is
represented schematically in Figure 1 and can be expressed algebraically as
X1 = X′1 , X2 = X′2 , X3
- 144 -
for which APQ is given by
- 145 -
RQAPQ
HP =
RJJQ00
1
0
1
0
−1
0
0 HJJP
. (23)
Thus, a coordinate transformation which reverses the direction of the X3 axis does not alter the
strain energy function. We will show that the number of material constants needed to describe
the behavior of a material with the above symmetry is reduced from 21 to 13.
For the symmetry under consideration, equation (17) must hold for the value of APQ given
in (23), but necessarily for other choices of APQ. Substituting (23) into (17) gives the relation-
ships
C1212 = C1212
C2313 = C2313
C3323 = −C3323
C2223 = −C2223
C1113 = −C1113
C1111 = C1111
C2312 = −C2312
C3313 = −C3313
C2213 = −C2213
C1112 = C1112
C1122 = C1122
C1313 = C1313
C3312 = C3312
C2212 = C2212
C2222 = C2222
C1133 = C1133
C1312 = −C1312
C2323 = C2323
C3333 = C3333
C2233 = C2233
C1123 = −C1123
which implies that
0 = C1123 = C1113 = C2223 = C2213 = C3323 = C3313 = C1312 = C2312 .
The remaining material constants from the set (13) are
C1111
C2222
C1122
C3333
C2233
C1133
C2323
0
0
0
C1313
C2313
0
0
0
C1212 .
0
0
C3312
C2212
C1112
(24)
The material properties of an elastic solid with material symmetry with respect to the X1 − X2
plane are characterized by the 13 constants in (24).
- 146 -
Case I(b) - Symmetry with respect to the X2−X3 plane
Consider now a solid that has material symmetry with respect to the X2−X3 plane. For such
a material, the mechanical behavior in the X1 direction is the same as the corresponding
behavior in the −X1 direction (denoted by X1 in Figure 2). The coordinate transformation for
which (17) holds, represented schematically in Figure 2, is given by
X1 = −X′1 , X2 = X′2 , X3 = X′3 . (25)
Since the material has symmetry with respect to the X2−X3 plane, a coordinate transformation
which reverses the direction of the X1 axis (such as that in (25)) will not alter the strain energy
function.
Case II - Orthotropy
A material which is symmetric with respect to two orthogonal planes is called orthotropic.iiiiiiiiii
Consider a homogenous elastic solid which has the material symmetry properties discussed in
I(a) as well as the symmetry properties discussed in I(b) (i.e. a material which is symmetric with
respect to both the X1−X2 plane and the X2−X3 plane). We will show that the number of
material constants needed to describe the mechanical behavior of a material with the above sym-
metries is 9. For the coordinate transformation given in equation (25), APQ is given by
RQAPQ
HP =
RJJQ 0
0
−1
0
1
0
1
0
0 HJJP
.
Substituting this choice for APQ into (17), we see that the 13 constants in (24) are related by
C1111 = C1111 ,
C2222 = C2222 ,
C1122 = C1122 ,
C1313 = C1313 ,
C2323 = C2323 ,
C3333 = C3333 ,
C2233 = C2233 ,
C1133 = C1133 ,
C1212 = C1212 ,
C2313 = −C2313
C3312 = −C3312
C2212 = −C2212
C1112 = −C1112
- 147 -
so that
0 = C1112 = C2212 = C3312 = C2313 .
The remaining independent constants are
C1111
C2222
C1122
C3333
C2233
C1133
C2323
0
0
0
C1313
0
0
0
0
C1212 .
0
0
0
0
0
(26)
The material properties of a linear elastic solid with material symmetry with respect to the
X1−X2 plane and with respect to the X2−X3 plane are thus characterized by the 9 constants given
in (26). Notice that by comparing (17) and (26), one can see that a material with the two sym-
metries described above is also symmetric with respect to a third plane which is orthogonal with
respect to both the original planes, i.e. the X1−X3 plane.
Case III - Transverse isotropy
A homogeneous orthotropic solid that also has material symmetry in every direction lying
in a certain plane (say, the X1−X2 plane) is said to be transversely isotropic.iiiiiiiiiiiiiiiiii For such a material,
a coordinate transformation which alters the X1 and X2 axes by an arbitrary rotation about the X3
axis will not alter the strain energy function. This property implies that equation (17) must hold
for a transformation of the type (also represented schematically in Figure 3)
X′1 = X1 cos α + X2 sin α
X′2 = −X1 sin α + X2 cos α (27)
X′3 = X3 ,
where the angle α is arbitrary. We will show that the number of material constants needed to
describe the mechanical behavior of a material with the above symmetry is 5. The components
- 148 -
of the orthogonal tensor APQ corresponding to (27) are
RQAPQ
HP =
RJQ 0sin αcos α
0cos α
− sin α
100 HJP
. (28)
Substituting (28) and (26), since the material is also orthotropic, into (17) gives the relationships
C1111 = C2222 , C1133 = C2233 , C1313 = C2323
C1212 =21hh(C1111 − C1122) .
The remaining independent compenents are
C1111
C1111
C1122
C3333
C1133
C1133
C1313
0
0
0
C1313
0
0
0
0
21hh(C1111 − C1122) .
0
0
0
0
0
(29)
The material properties of a linear elastic solid which is transversely isotropic are thus character-
ized by the five constants given in (29).
Case IV - Isotropic material
A homogeneous elastic solid which has material symmetry in the reference configuration
with respect to all directions is said to be isotropic.iiiiiiii Thus, equation (17) will remain valid for any
proper orthogonal choice of APQ. We will now derive equation (18) using the symmetries which
characterize an isotropic material. We will thus show that only two material constants are
needed to describe the mechanical behavior of a linear elastic isotropic solid.
An isotropic material can be considered as a material transversely isotropic in three mutu-
ally orthogonal planes (such as the X1−X2, X1−X3, and X2−X3 planes). Thus, a coordinate
transformation which alters the X2 and X3 axes by an arbitrary rotation about the X1 axis will not
- 149 -
alter the strain energy function (see Figure 4a). The same is true with respect to an arbitrary
rotation of the X1 and X3 axes about the X2 axes (see Figure 4b).
Thus, equation (17) must remain valid under the transformations
X′1 = X1
X′2 = X2 cos φ + X3 sin φ (30)
X′3 = −X2 sin φ + X3 cos φ
and
X′1 = X1 cos β − X3 sin βX′2 = X2 (31)
X′3 = X3 cos β + X1 sin β ,
where φ and β are arbitrary angles, in addition to the transformation (27). For the coordinate
transformation in (30), APQ is given by
RQAPQ
HP =
RJJQ00
1
sin φcos φ
0
cos φ−sin φ
0 HJJP
(32)
and for the coordinate transformation in (31), APQ is given by
RQAPQ
HP =
RJJQ−sin β
0
cos β
0
1
0
cos β0
sin β HJJP
. (33)
>From (17), (29), (32), and (33), we can show the relationships
C1122 = C1133 , C1111 = C3333
C1212 = C1313 =21hh(C1111 − C1122) .
The remaining independent constants are given by
C1111
C1111
C1122
C1111
C1122
C1122
µ
0
0
0
µ0
0
0
0
µ0
0
0
0
0
(34)
- 150 -
where µ =21hh(C1111 − C1122). Letting λ = C1122, equation (34) may be expressed as
λ + 2µλ + 2µ
λ
λ + 2µλλ
µ0
0
0
µ0
0
0
0
µ .
0
0
0
0
0
(35)
An equivalent way of writing (35) is given simply
CABMN = λδABδMN + µ(δAMδBN + δANδBM) . (36)
Notice that (36) is the same as equation (18) obtained previously.
- 151 -
Appendix L: Some Results from Linear Algebra
The object of this appendix is to introduce the concept of a tensor in a form which is espe-
cially useful in continuum mechanics. Thus, after a rapid review and summary of basic results
from linear algebra, a tensor is regarded as a linear transformation. Many aspects of the algebra
and the calculus of tensors are subsequently discussed in coordinate-free form and frequently are
represented with reference to rectangular Cartesian basis vectors.
1. Sets.
We use the symbol X to represent a set, class, collection or family which contains elements
or objects x1,x2,.... We write this as
X = x1,x2,.... (1.1)
The element x is a member of X, or belongs to X, or is in X, and this is denoted by
x ∈ X. (1.2)
Let X and Y be two sets. The set X is a subset of Y if every x in X is also in Y. This rela-
tion is designated by
X ⊆ Y. (1.3)
Also Y is said to include or contain X, and we write Y ⊇ X. The empty set is contained in every
set and is designated by ∅. If
X ⊆ Y and Y ⊆ X, (1.4)
then the set Y and X are said to coincide or be equal and we write
Y = X. (1.5)
- 152 -
If X is a subset of Y but without X being equal to Y, i.e., X ⊂ Y, then X is a proper subset
of Y and we write1
X ⊂ Y , Y ⊃ X. (1.6)
It should be clear that X ⊂ Y means that (i) if x ∈ X, then x ∈ Y and that (ii) there exists a y ∈
Y such that y ∈ X.
The union of two sets X and Y comprises all the elements of the two sets and is denoted by
X ∪ Y. (1.7)
The intersection of two sets consists only of those elements which belong to both X and Y and is
denoted by
X ∩ Y. (1.8)
The Cartesian product of two sets X and Y is the set
X × Y = (x,y) : x ∈ X, y ∈ Y (1.9)
of ordered pairs. For example, if X = x1,x2,x3 and Y = y1,y2, then X × Y =
(x1,y1),(x1,y2),(x2,y1),(x2,y2),(x3,y1),(x3,y2) .
For completeness, we introduce now the following definition of a function: A function f
from the set X to the set Y is a rule which assigns to every element of X a unique element of Y.
It is often displayed as
f : X → Y , x |→ y . (1.10)
hhhhhhhhhhhhhhh1 Our use of the symbols ⊆ and ⊂ is analogous to the use of the symbols ≤ and < to denote weak and strict ine-
qualities.
- 153 -
The first of (1.10) indicates that the set X is transformed into Y, while the second of (1.10) indi-
cates that the element x goes to the element y. The function f, or the rule defined by (1.10)1, is
sometimes called an operation or a transformation or a mapping. This definition of a function
rules out multivalued functions.
- 154 -
2. Vector spaces.
Consider a set V of arbitrary elements u,v,w,...,etc., and admit the following two operations
labeled as (1) and (2):
(1) Let there exist an operation, denoted by "+", which to every pair u,v assigns a unique
element u + v in V and which has the properties:
(A1) u + v = v + u (commutative property).
(A2) u + (v + w) = (u + v) + w (associative property).
(A3) There exists a zero vector o such that
v + o = v .
(A4) For every vector v there is a corresponding negative vector (−v)
such that
v + (−v) = o.
It is customary to write v + (-w) = v - w.
(2) Let there exist another operation, indicated by placing a juxtaposition of an element of
V and a real number α , which to every element v assigns a unique element αv in V and which
has the following properties:
(S1) 1v = v.
(S2) α(βv) = (αβ)v (associative property).
(S3) (α+β)v = αv + βv (distributive property for scalar addition).
(S4) α(v+w) = αv + αw (distributive property for vector addition).
- 155 -
We then say that V is a vector space over the field of real numbers and that u,v,w,... are
vectors in V. The first operation defined under (1) above is callled vector addition with elements
u + v called the sum of u and v. The second operation defined under (2) is called multiplication
of vectors by a real number.
All the usual algebraic results can be derived from the above axioms. For example, a vec-
tor equation of the form αu = o has the solution α = 0 or u = o.
Vector sub-spaces. A sub-space of a vector space V is any nonempty subset U of V which
is such that if u and v belong to U and α is any real number, then the vectors u + v and αv also
belong to U.
- 156 -
3. n-Dimensional vector spaces.
Let v1,v2,..,vp be any p vectors in a vector space V, p being some integer. The sumi=1Σp
αi vi
is called a linear combination of the p vectors vi with coefficients αi as real numbers. These vec-
tors are said to form a linearly independent set of order p if the only coefficients that satisfy the
equation
α1v1 + α2v2 + . . . + αpvp = o (3.1)
are α1 = α2 = α3 = . . . = αp = 0. A linearly dependent set is one which is not linearly
independent.
Note that every subset of vectors containing the zero vector is linearly dependent.
α1v1 + α2v2 + . . . + 1 o + αpvp = 0.
Consider the set of all systems of linearly independent vectors in the vector space V. There
are two possibilities: Either (1) there exist linearly independent systems of arbitrarily large order
p; then the vector space is said to be an infinite-dimensional space, or (2) the order of the linearly
independent sytstem is bounded. In the second case, there exists an integer n such that the order
p ≤ n and there exist linearly independent systems of order n but not of n + 1. The vector space
V is said to be a finite-dimensional vector space. The number n is termed the dimension of the
vector space and we shall use V n to designate finite n-dimensional vector space; and henceforth
in this Appendix we shall be concerned with finite dimensional vector spaces only. Let
g1,g2,..,gn be any such system of order n in a finite dimensional vector space V n ; it will be
called a basis of V n . Thus, we have the following definition:
A basis in a vector space V n is any linearly independent set of n vectors.
Let v be any vector in V n. The set of n + 1 vectors v,g1,g2,..,gn is necessarily linearly
dependent, so there exists n + 1 numbers λ,α1,α2,..,αn not all zero such that
- 157 -
λv + α1g1 + α2g2 + . . . + αngn = o. (3.2)
If v = o, then αigi = o, (no sum on i), and hence αi = 0. For v ≠ o, the number λ must
be different from zero, for otherwise the system gi (i = 1,2,..,n) will not be linearly independent.
Equation (3.2) can, therefore, be solved for v, and there exist numbers v1,v2,...,vn such that
v = v1g1 + v2g2 + . . . + vngn . (3.3)
Thus, the vector v is expressible as a linear combination of the gi . This combination is unique;
for otherwise, there would exist another, and their difference would constitute a linear combina-
tion of the gi equal to the null vector o and with coefficients not all zero. The numbers
(v1,v2,..,vn) are called the components of v with respect to the basis g1,g2,..,gn . It is con-
venient to replace (3.3) by the shorter notation
v = vkgk (3.4)
in which the summation convention is used. Whenever an index appears twice in the same term,
a summation is implied over all terms by letting that index assume all its possible values, unless
the contrary is stated. Normally the convention applies to an index which appears once as a sub-
script and once as a superscript; but, as will become apparent later, in a special case it will
suffice to adopt the convention for repeated subscripts. It is convenient to recall here a theorem
the proof of which can be easily found in a book of linear algebra2. A statement of the theorem
is as follows:
For a set of vectors to constitute a basis in V n it is necessary and sufficient that every vector
in V n can be expressed in one, and only one, way as a linear combination of the vectors of that
set.
hhhhhhhhhhhhhhh2 See a standard book on linear algebra.
- 158 -
4. Euclidean vector spaces.
Consider a vector space of dimension n defined over the field of real numbers. Suppose
there is an operation, denoted by " . ", which to every pair of vectors u and v assigns a real
number, denoted by u . v, and which has the following properties:
(I1) u . v = v . u .
(I2) u . (v+w) = u . v + u . w .
(I3) (αu) . v = u . (αv) = α(u . v) .
(I4) u . u ≥ 0 and u . u = 0 => u = o .
The above operation (known as a rule of composition) is called the dot product or the scalar pro-
duct or the inner product of two vectors. A vector space V n obeying the rules of composition
(I1) to (I4) is a real inner product space and is called a Euclidean vector space3 E n .
The magnitude or the norm of the vector v is defined by
c c v c c = (v.v)1⁄2 . (4.1)
If c c v c c = 1, the vector is said to be normalized. By (I4) c c v c c 2 is always positive if v ≠
o and it vanishes when v = o , so that its square root is real.
The norms of two vectors and the magnitude of their scalar product satisfy the Schwarz ine-
quality
hhhhhhhhhhhhhhh3 This terminology stems from the fact that once a scalar product is admitted, all the standard theorems of
Euclid’s geometry can be established by an appeal to the properties of an inner product space.
- 159 -
c u . v c ≤ c c u c c c c v c c . (4.2)
To prove this, consider the vector αu + v where α is an arbitrary real number. Then,
c c (αu + v) c c 2 = α2 c c u c c 2 + 2αu . v + c c v c c 2 .
The left-hand side of the last expression is positive or zero for all real α and this implies
(u . v)2 ≤ c c u c c 2 c c v c c 2 ,
which is equivalent to (4.2). In view of (4.2), we define the angle between two nonzero vectors
u and v by
cos θ =c c u c c c c v c c
u . vhhhhhhhhh , (4.3)
where 0 ≤ θ ≤ π.
Vectors u and v are said to be orthogonal if
u . v = 0 . (4.4)
A set of vectors in E n is said to be orthonormal if all the vectors in the set are normalized
and mutually orthogonal. Let ei be a system of orthonormal basis in E n . It follows that ei must
satisfy the conditions
ei. ek = δik , (4.5)
where δik (i,k = 1,2,...,n) is the Kronecker delta defined by
δik =IKL1
0
if
if
i
i
=
≠
k
k
.
,(4.6)
Every orthonormal set is necessarily linearly independent. An infinity of bases exist which are
orthonormal; and in this appendix a typical orthonormal basis will be denoted by ei . Any vector
- 160 -
v in E n, when referred to the basis ei, can be expressed in the form
v =i=1Σn
viei . (4.7)
Then, by taking the scalar product of both sides of (4.7) with ek , we have
vk = v . ek . (4.8)
The numbers vk are the components of v with respect to the orthonormal basis ek and the square
of the magnitude of v is
c c v c c 2 = v . v =i=1Σn
vivi . (4.9)
- 161 -
5. Points.
We denote points by boldface letters x, y, z, etc. Given any two points x and y , we associ-
ate with them a point difference y − x which is a vector v in E n and which we denote by
v = y − x (5.1)
satisfying the following rules:
(P1) (y − x) + (x − z) = (y − z) .
(P2) Given any point x and any vector v, there is a unique point y such
that y − x = v .
The point y determined in rule (P2) is denoted by
y = x + v = v + x . (5.2)
In view of (5.1) and (5.2), it is meaningful to speak of the difference of two points which is
a vector, and of the sum of a point and a vector , which is a point. In particular,
x − x = o (5.3)
holds for every point x .
The distance between two points x,y is
c x − y c = (x − y) . (x − y)1/2 . (5.4)
- 162 -
The set of all points associated with a Euclidean vector space E n is called Euclidean Point
Space.4
hhhhhhhhhhhhhhh4 Euclidean vector and point spaces are abstract concepts which at this point in our discussion are unrelated to or-
dinary vectors and points of elementary geometry. Geometrical representations of these abstract concepts will bediscussed in Section 6.
- 163 -
6. Geometric space.
Consider the space of elementary geometry. This space is composed of elements P,Q,R,
called points. Choose O to be a reference point or origin. The directed line segments
OP→
, OQ→
,... may be associated with points x,y,... of Euclidean point space (Section 5) and the
directed line segments PQ→
, QR→
may be associated with vectors u,v,... in Euclidean vector space
(Section 4). Thus we set
x = OP→
, y = OQ→
, z = OR→
,... (6.1)
and
u = PQ→
, v = QR→
,... o = PP→
. (6.2)
The properties (A1)−(A4) of abstract vectors and the properties (P1),(P2) of points have their
immediate geometrical representation in our geometric space. Thus, the class of all line seg-
ments with the same direction and length as PQ→
, called a geometrical vector or free vector,
represent vectors, and directed segments such as OP→
, called position vectors, represent points.
For example, we have the properties
PQ→ = −QP
→, PQ
→ + QR→ = PR
→.
Next, the properties (S1)−(S4) have their geometric representation in our space of elemen-
tary geometry. Multiplication of a vector by a scalar corresponds to multiplication of a line ele-
ment PQ→
by a scalar α and is represented by a line element of length c α c c PQ c , where c PQ c is
the length of PQ→
in the same or opposite direction as PQ→
according to whether α is positive or
negative.
The rules (I1)−(I4) of Section 4 have their representation in the geometrical space. Recal-
ling also (4.3), these rules are represented by the scalar product of two geometrical vectors as
PQ→ . RS
→ = c PQ c c RS c cos θ , (6.3)
- 164 -
where θ is the angle between PQ→
and RS→
. Thus, geometrical vectors give a geometrical
representation of a Euclidean point space with its associated Euclidean vector space. Geometri-
cal space has three dimensions.
Given a geometrical space, we select an origin O and a (constant) basis fi of the associ-
ated vector space. These are said to constitute a coordinate system consisting of origin and a
basis (O,fi) for our geometrical space. Any point x referred to the basis fi can be expressed in
terms of its components xh
i by the formula
x =i=1Σ3
xhifi , (6.4)
where now summation is over i = 1,2,3. The components xhi are called rectilinear coordinates of
x. If (fi) is identified with a constant orthonormal basis ei, then referred to ei the point x can be
represented as
x =i=1Σ3
xiei (6.5)
and xi are called Cartesian rectilinear coordinates of x.
We may select a different basis gi (say) at each point x in our geometric space so that
gi = gi(x). This situation arises when we consider curvilinear coordinates in geometric space.
- 165 -
7. Indicial notation.
- 166 -
8. Linear transformation. Tensors.
A linear transformation or tensor T is an operation which assigns to each vector v in V n
another vector Tv in V m such that5
T(v+w) = Tv + Tw , T(αv) = αTv (8.1)
for all v,w in V n. Thus, we have T: V n → V m .
We denote the set of all possible tensors which transform vectors in V n linearly into vec-
tors in V m by L(V n,V m) and we define the sums of two tensors and the scalar multiples of ten-
sors by
(T+S)v = Tv + Sv , (αT)v = α(Tv) . (8.2)
The transformations T + S and αT defined in (8.2) obey the rules (8.1), i.e., they are tensors.
Also the sums and scalar multiples defined in L(V n,V m) obey the rules (A1)−(A4) and (S1)−(S4).
Hence the set of all tensors constitutes a vector space. The zero element in L(V n,V m), i.e., the
element which is to be substituted for o in rules (A3) and (A4) is the transformation which
assigns the zero vector to every vector. We call this transformation the zero tensor and denote it
by O. In other words, the zero tensor O is such that
Ov = o (8.3)
for all vectors v.
The identity tensor can only be defined for transformations from V n to V n. Thus, when the
dimension m = n, we define the identity or the unit tensor I by
Iv = v (8.4)
hhhhhhhhhhhhhhh5 This special definition is particularly useful in the present Appendix. A more general definition of a tensor is
given in Appendix C.
- 167 -
for all vectors v in V n.
We associate with Euclidean vector spaces E n and E m a vector space of dimensions nm
denoted by E n ×b E m called the tensor product space of E n and E m, and defined in the follow-
ing way: If a ∈ E n and b ∈ E m, then a ×b b is an element of E n ×b E m such that
(a ×b b)v = a(b . v) (8.5)
for all vectors v ∈ E m. It is seen that a ×b b, which sometimes is denoted as ab, obeys the rules
(8.1) and hence is a tensor, i.e., it assigns to each vector v ∈ E m the vector a(b . v) ∈ E n.
Sums and scalar multiples of such tensors can then be calculated by the rules laid down for ten-
sors in (8.2). Not all tensors generated in this way can be expressed as an element of space of
the form a ×b b. The zero tensor in E n ×b E n is defined by the tensor o ×b o.
Consider now the tensor product space E n ×b E n of dimension n2. Let ei be an orthonormal
basis in Euclidean vector space E n. Then,
ei ×b ek (8.6)
is a basis in E n ×b E n. To prove this, suppose αik are a set of n2 numbers such that
αikei ×b ek = O .
Then,
(αikei ×b ek)er = Oer = o .
With the help of (8.5) this last equation becomes
αikeiδkr = αirei = o .
Since ei is a basis in E n it follows that
- 168 -
αik = 0 .
Hence ei ×b ek are linearly independent and form a basis in E n ×b E n. When u and v are vectors
in E n so that
u = uiei , v = viei ,
then
uv = u ×b v = uivkei ×b ek . (8.7)
Let T be a tensor in L(E n,E n) and let tik denote the components of the vector Tek with respect
to the basis ei in E n so that
Tek = tikei , tik = Tek. ei . (8.8)
Associated with each tensor T in L(E n,E n) we have the ordered array of n2 scalars tik which
combine by the rules (A1)−(A4) and (S1)−(S4) for vector spaces. Hence, L(E n,E n) is a vector
space of dimension n2. But, ei ×b ek is a linear transformation in the space L(E n,E n) and hence
also forms a basis in L(E n,E n) which is identical to E n ×b E n. Moreover, with the use of (8.8),
for any vector v ∈ E n we have
(T−tikei ×b ek)v = Tekvk − tikeivk = o .
Hence, we conclude that
T = tikei ×b ek . (8.9)
The n2 numbers tik in (8.9) are called the components (or Cartesian components) of T with
respect to the basis ei ×b ek .
Similarly, we write
I = δikei ×b ek (8.10)
- 169 -
so that
Iv = v (8.11)
for all vectors v in E n. The tensor I with components δik is the unit tensor in E n ×b E n or
L(E n,E n).
The system of numbers tik in (8.8) is usually regarded as an array of numbers, a matrix,
having n rows and n columns and denoted by [tik] or tik.
Let T be a tensor in E n ×b E n. The transpose of a tensor T is a tensor TT defined by
w . Tv = v . TTw (8.12)
for all vectors v and w in E n. The transpose TT satisfies (8.1) and is a tensor, and the opera-
tion which associates TT with T is called transposition. From (8.12) it may be shown that
(TT)T = T , (T+S)T = TT + ST , (αT)T = αTT , (8.13)
so that transposition is a linear operation. Further, it follows from (8.9) and (8.12) that
[tik] = ei. Tek = ek
. TTei = [tik]T ,
where [tik]T are components of the tensor TT.
A tensor T is said to be symmetric if
TT = T . (8.14)
Also, by (8.9) and (8.14) we have
tki = tik . (8.15)
Also, from (8.12) and (8.14) it follows that if T is symmetric, then
w . Tv = v . Tw (8.16)
- 170 -
for every v,w in E n and conversely.
A tensor T is said to be skew or anti-symmetric (skew- symmetric) if
TT = −T (8.17)
or
tki = −tik . (8.18)
Every tensor T in E n ×b E n can be written as
T = TS + TA , (8.19)
where
TS =21hh (T+TT) − (TS)T , TA =
21hh (T−TT) = −(TA)T . (8.20)
- 171 -
9. Multiplication of tensors.
Let T and S be two tensors in L(V n,V n). The product TS of two tensors T and S is the
composition of T and S , i.e., TS defined by the requirement that
(TS)v = T(Sv) (9.1)
hold for all vectors v in V n. It is seen that TS satisfies (8.1) and is a tensor in L(V n,V n). The
following rules may be established
(M1) (TS)R = T(SR) ,
(M2) T(R+S) = TR + TS ,
(M3) (R+S)T = RT+ST ,
(M4) α(TS) = (αT)S = T(αS) ,
(M5) IT = TI = T ,
where T,S,R are tensors in L(V n,V n).
For tensors T,S in L(E n,E n) it follows from (6.9) and (8.1) that
(TS)v = T(smkvkem) = timsmkvkei
for all vectors v ∈ E n so that
TS = timsmkei ×b ek (9.2)
and timsmk are the components of TS with respect to the basis ei ×b ej. If tik and sik are
matrix representations of the tensors T and S, respectively, then
tiksik = timsmk . (9.3)
- 172 -
Note that
(TS)T = STTT . (9.4)
The commutative law is not, in general, satisfied, i.e., TS is not the same as ST.
The truth of the following formulae can be readily verified:
(a ×b b)T = b ×b a ,
T(a ×b b) = Ta ×b b ,
(9.5)
a ×b (TTb) = (a ×b b)T ,
(a ×b b)(c ×b d) = (b . c)a ×b d .
The tensor T is invertible if, for every choice of w, the equation w = Tv can be solved
uniquely for the vector v. If T is invertible, then v is unique and we write v = T−1w. The
transformation T−1 obeys the rules (8.1) and is a tensor called the inverse of T. If the inverse
T−1 exists, then
(TT−1−I)w = o , (T−1T−I)v = o
for every choice of w and every choice of v so that
TT−1 = T−1T = I , (T−1)−1 = T . (9.6)
If S and T are invertible tensors in L(V n,V n) and if w = (TS)v, it follows that
Sv = T−1w , v = S−1T−1w .
Hence, the product TS is invertible and
(TS)−1 = S−1T−1 . (9.7)
- 173 -
If the only solution of the equation Tv = o is v = o, then T is nonsingular; otherwise it
is singular. If T is invertible, then T is nonsingular. It can be shown that a tensor T is invertible
or nonsingular if and only if det T ≠ 0.
A tensor Q is orthogonal if it is invertible and if the inverse coincides with its transpose,
i.e.,
Q−1 = QT or QQT = QTQ = I . (9.8)
A tensor Q is orthogonal if and only if it preserves inner products in the sense that
(Qu) . (Qv) = u . v (9.9)
for every vector u,v in E n.
The tensor T is positive semidefinite if it is symmetric and if v . Tv ≥ o for all vectors v in
E n. If v . Tv > o when v ≠ o, then T is positive definite.
An operation tr which assigns to each tensor T in E n ×b E n a number tr T is a trace if
tr(S+T) = tr S + tr T , tr(αT) = α tr T , (9.10)
tr a ×b b = a . b , (9.11)
for all vectors a,b in E n and all scalars α. The tr operation is a linear function and the follow-
ing consequences are clear:
tr T = tikei. ek = tii . (9.12)
Also,
tr TT = tr T , tr(TS) = tr(ST) ,
(9.13)
tr I = n , tr TS = tr T , tr TA = 0 ,
- 174 -
where n in (9.12)3 is the dimension of the space.
If one defines the inner product of two tensors T,S to be the number
T . S = tr(TST) , (9.14)
then the rules (I1)−(I5) are satisfied. Hence with tr(STT) as the inner product, the space of all
tensors is an inner product space. We define the magnitude of T as
| T | = √ddddddtr(TTT) . (9.15)
- 175 -
10. Change of basis.
Let ei and eh
i be two arbitrary orthonormal bases in E n. Each vector in one basis may be
expressed in terms of those of the other basis by a nonsingular transformation. Thus
eh
i = Aei = akiek ,
(10.1)
ei = Ahh
eh
i = ah
kieh
k , Ahh
= A−1 ,
where
eh
k. Aei = δik = ei
. Ahh
eh
k ,
(10.2)
ah
ki = ei. eh
k = aik , Ahh
= AT .
Hence, A−1 = AT and A is orthogonal.
If v is an arbitary vector and
v = viei = vh
ieh
i , (10.3)
then the components of v in the two bases ei and eh
i are related by the equations
vh
i = akivk , vi = aikvh
k . (10.4)
Also
v . v = vivi = vh
ivh
i , u . v = uivi = uh
ivh
i , (10.5)
and the scalar product has the same form in terms of the components ui,vi as it has in terms of
the components uh
i,vh
i and is called a scalar invariant.
Again, if T is an arbitrary tensor in E n ×b E n and
T = tikei ×b ek = th
ikeh
i ×b eh
k , (10.6)
- 176 -
then the components of T in the two bases ei and eh
i are related by
th
ik = ariasktrs , tik = airaksth
rs . (10.7)
If T and S are two tensors in E n ×b E n and
S = sikei ×b ek = sh
ikeh
i ×b eh
k ,
then
T .S = tirsir = th
irsh
ir ,
(10.8)
T . T = tirtir = th
irth
ir
are scalar invariant forms.
- 177 -
11. Point, vector and tensor functions.
The distance between two points x,y is defined in (6.4). The distance vanishes if and only
if x = y. This idea of distance can be used to define limits, convergence, continuity, etc. For
example, we say that
n→∞lim xn = x if
n→∞lim | xn−x | = O . (11.1)
Suppose z(t) is a function of a real variable t called a point function. The derivative z.(t),
if it exists, on an open subset of the real numbers is defined by
z.(t) =
dtdzhhh =
∆t→∞lim
∆tz(t+∆t) − z(t)hhhhhhhhhhhh . (11.2)
Such a derivative is a vector-valued function.
In a similar way we may define limits, derivatives, etc., of vector- and tensor-valued func-
tions v,T of t using the appropriate definitions (1.2) and (9.15) for distance. The derivative of a
vector-valued function of t is a vector function and the derivative of a tensor-valued function is a
tensor function. The usual rules of differential calculus are easily extended so as to apply to
point, vector or tensor functions. For example, if T = T(t) and S = S(t) are tensors, then
TShhh.
= T.S + TS
., (11.3)
where, of course, the correct order must be maintained for the tensors.
If Q = Q(t) is an orthogonal tensor function, then from (9.8) and (11.3),
Q.
QT + QQThhh.
= O . (11.4)
Since Q.
T = QThhh
it follows that
Ω = Q.
QT = − Q Q.
T (11.5)
- 178 -
is a skew tensor.
If φ(x) is a scalar function of x defined on some open region U of Euclidean point space,
then φ is differentiable on U if there is a vector field w(x) on U such that
x→oxlim
| x−ox |
| φ(x) − φ(ox) − w(ox) . (x − ox) |hhhhhhhhhhhhhhhhhhhhhhhhhhh = 0 (11.6)
for everyox in U. If this is so, w is unique. We call it the gradient of φ and write
w = grad φ(x) =∂x∂φhhh . (11.7)
Alternatively, the gradient of φ(x) may be defined by means of
β→0lim
βφ(x+βu) − φ(x)hhhhhhhhhhhhh =
dβdhhh φ(x + βu) c
c β=0
=∂x∂φhhh . u = w . u (11.8)
for all vectors u ∈ E n. It may be shown that this definition of gradient is equivalent to that
given in (11.6).
Sometimes we use the symbol ∇ to denote the gradient operator:
∇ =∂x∂hhh = er ∂xr
∂hhhh (11.9)
if x = xrer and er is an orthonormal basis. From (11.6) it follows that
w = grad φ(x) =∂x∂φhhh = ∇φ = ei ∂xi
∂φhhhh . (11.10)
If v(x) is a vector function of points x in some open region U of Euclidean space, then
v(x) is said to be a vector field. The gradient of a vector field v(x) is a tensor field T(x)
defined by
- 179 -
x→oxlim
| x−ox |
| v(x) − v(ox) − T(ox)(x−ox) |hhhhhhhhhhhhhhhhhhhhhhhhhh = o (11.11)
for every ox in U. We write this as
T = grad v(x) =∂x∂vhhh . (11.12)
If v = viei, then it follows from (11.12) that
T =∂xk
∂vihhhhei ×b ek (11.13)
so that∂xk
∂vihhhh are the components of the grad v with respect to the orthonormal basis ei. It can
be shown that
grad v(x)Tc = gradc.v(x) (11.14)
for every constant vector c. Equation (11.14) could be taken as the defining equation for grad
v(x).
The divergence of the vector field v(x) is a scalar field defined by
div v(x) = trgrad v(x) = ∇ . v = ei.
∂xi
∂vhhhh =∂xi
∂vihhhh . (11.15)
The divergence of a tensor field T(x) is the vector field div T(x) for which
c . div T(x) = divTT(x)c (11.16)
for every constant vector c. When
T = tijei ×b ej , (11.17)
- 180 -
it follows that
div T(x) =∂xj
∂tijhhhh ei . (11.18)
If P is a bounded region whose boundary ∂P is sufficiently well-behaved, then by applica-
tion of the divergence theorem (Green’s theorem) to a vector v and a tensor T we have
∫P
div v(x)dV(x) = ∫∂P
v(x) . n(x)dA(x) , (11.19)
∫P
div T(x)dV(x) = ∫∂P
T(x)n(x)dA(x) , (11.20)
where n is the outward unit normal to ∂P, the integrations are over the volume P and the boun-
dary surface ∂P and dV and dA represent elements of volume and area, respectively. In com-
ponent form, (11.19) and (11.20) read:
∫P
vi,i dv = ∫∂P
vi ni da , (11.21)
∫P
tij,j dv = ∫∂P
tij nj da . (11.22)
- 181 -
12. Vector product. Axial vectors. The curl operator.
Consider a Euclidean vector space E3 of three dimensions. If u,v,w are any vectors in E3 ,
the vector product of u and v is a vector in E3 denoted by u × v and defined by the properties:
(V1) w.(u×v) = u.(v×w) = v.(w×u) and is a scalar denoted by
[uvw] and called the scalar triple product ,
(V2) u × v = −v × u ,
(V3) | u×v | = | u | | v | sin θ ,
where θ is defined in (4.3). From (V1) and (V2), it follows that
u × u = o , u.(u×v) = 0 , v.(u×v) = 0 (12.1)
so that u × v is orthogonal to both u and v. It can be shown that three nonzero vectors u,v,w
are linearly independent if and only if their scalar triple product [u,v,w] = u.(v×w) = (u×v).w.
The following properties may be deduced from the definitions:
(αu) × v = α(u×v) = u × (αv) ,
u × (v+w) = u × v + u × w ,
(u+v) × w = u × w + v × w , (12.2)
u × (v×w) = v(u.w) − w(u.v) = (v ×b w − w ×b v)u ,
[u + x,v,w] = [uvw] + [xvw] .
The above definition of vector product does not restrict ei to be either right-handed or left-
handed basis.
- 182 -
Let e1,e2,e3 be an orthonormal system. Then, in view of (V3) and (12.1),
e1 × e2 = ± e3 . (12.3)
A right-handed orthonormal sytem is one for which
e1 × e2 = e3 (12.4)
and a left-handed orthonormal system corresponds to the choice of the - sign in (12.3). For a
right-handed orthonormal system it follows that
[e1e2e3] = 1 , ei × ej = eijkek , eijk = [eiejek] , (12.5)
where the components eijk are known as the permutation symbol or the components of the alter-
nating tensor. Also, for any two vectors
u = uiei , v = vjej ,
(12.6)
u × v = eijkuivjek .
Let TA be a skew tensor in E3 so that
TA = tikei ×b ek =21hh tik[ei ×b ek − ek ×b ei] . (12.7)
Hence, if u is an arbitrary vector in E3
TAu =21hh tik[ei(u.ek) − ek(u.ei)] =
21hh u × [tikei × ek] , (12.8)
if we use (12.2)4. Hence
TAu = tA × u , (12.9)
- 183 -
where the vector
tA =21hh tkiei × ek . (12.10)
Let ei be a right-handed orthonormal basis. Then, referred to ei, TA reads
TA = tijei ×b ej (12.11)
and the axial vector tA with components tkA assumes the form
tA =21hh tjieijkek = tk
A ek . (12.12)
Alternatively
tji = eijktkA . (12.13)
If
TA =∂x∂vhhh − (
∂x∂vhhh)T , (12.14)
the corresponding axial vector tA is called the curl of v and written as
tA = curl v . (12.15)
In a right-handed orthonormal system
curl v = eijk ∂xi
∂vjhhhh ek . (12.16)
- 184 -
References for Appendix L
Gel’fand, I. M. 1961 Lectures on linear algebra (translated by A. Shenitzer from the 2nd Rus-sian edition). Dover, New York.
Halmos, P. R. 1974 Finite dimensional vector spaces. Springer-Verlag, New York.
Hoffman, K. & Kunze, R. 1971 Linear algebra (2nd edition). Prentice-Hall, Inc., EnglewoodCliffs, New Jersey.
Stecher, M. 1988 Linear algebra. Harper & Row, New York.
